What is the theory of relativity?

[groom03]

I'm trying to understand the theory of relativity can someone tell me (in English!) if I'm close to correct.

The theory of relativity says that the closer you get to the speed of light the faster time goes (i think someone said it was impossible to get to the speed of light)

e.g two people got to somewhere really far away
person A is at 0.99 the speed of light and person B isn't

the journey for person B takes 100 years but for person A only a short time has passed.

Does that make any sense?

 

[ZapperZ]

I think you got "time dilation" pretty wrong.

You may want to start from scratch and read up a bit on the basics of Special Relativity. Try here:

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relcon.html

Zz.

 

[groom03]

so in english, what is the theory of relativity?

 

[russ_watters]

In plain English, Special Relativity says:

1. The laws of the universe in a non-accelerating reference frame are the same everywhere.
2. The speed of light is constant.

But that doesn't really tell you how time dilation works...

 

[groom03]

I've think I've got the idea of time dilation, time is perceived as constant for everyone but time could be going for faster for someone than me but that person would still think that time is passing normally.

so am i right about the idea of time sppeding up for me as i got closer to the speed of light?

 

[JesseM]

I've think I've got the idea of time dilation, time is perceived as constant for everyone but time could be going for faster for someone than me but that person would still think that time is passing normally.

so am i right about the idea of time sppeding up for me as i got closer to the speed of light?

In relativity there is no objective notion of speed, you can only talk about your speed relative to some other observer. If I measure you to be moving close to the speed of light in my rest frame, then I will measure your clock to be ticking slower than mine. But as long as both of us are moving inertially (constant speed and direction, no acceleration) the situation is symmetrical--you measure me to be moving close to the speed of light in your rest frame, and you measure my clock to be ticking slower than your own. If one of us accelerates to catch up with the other, though, this symmetry is broken, and whichever one accelerated will be found to have aged less between the time we first passed each other and the time we reunited later. You might want to read some information on the twin paradox to learn more about this.

 

[Hootenanny]

so am i right about the idea of time sppeding up for me as i got closer to the speed of light?

No, your (received) time period would be less than that measure by a 'stationary observer', hence, time would 'slow down' for you. I suggest you re-read the link which Zappers gave you and apply the Lorentz transformations (or more particularly the time dilation formula) to your situation.

Edit: JM got there before me

 

[DaveC426913]

You've got the time dliation backwards.

The short part that you are looking for, wrt time dilation, is that if you traveled at close to the speed of light for a long time, time outside your spaceship would appear to march by faster. When you arrive at your destination, decades or centuires may have passed everywhere but aboard your ship.

You'll need to read up to understand it better, it's not simple.

 

[JesseM]

You've got the time dliation backwards.

The short part that you are looking for, wrt time dilation, is that if you traveled at close to the speed of light for a long time, time outside your spaceship would appear to march by faster. When you arrive at your destination, decades or centuires may have passed everywhere but aboard your ship.

That statement is somewhat questionable--it depends what you mean by "time outside your spaceship would appear to march by faster". If you're traveling from Earth to Alpha Centauri, then in the rest frame of your ship, clocks on Earth and Alpha Centauri are ticking slower, not faster, since after all in your rest frame it is they that are moving at relativistic speeds while you are at rest. However, in your frame the clock at Alpha Centauri is considerably ahead of the one on Earth (that's the relativity of simultaneity--the two clocks are synchronized in the rest frame of the Earth and Alpha Centauri), so that explains why the difference of (time at Alpha Centauri's clock when you arrive) - (time on Earth's clock when you depart) will be greater than the time it took you to get from Earth to Alpha Centauri according to your own clocks.

On the other hand, if you're not talking about what's happening in the ship-observer's frame, but rather what the ship-observer sees using light-signals, then due to the Doppler shift, you'll see clocks at Alpha Centauri ticking faster than your own as you move towards it, while you'll see clocks at Earth ticking slower than your own as you move away from it (slowed down by a greater factor than just the time dilation factor). The issue of what you see vs. what you measure in your rest frame was discussed at length on this thread.

 

[jtbell]

It's best not to think in terms of the speed that you are traveling at, because (a) it depends on what you're measuring the speed with respect to, and (b) in your own reference frame, you yourself are always stationary!

It's better to think in terms of the speed(s) that other objects are traveling at, in your reference frame, when analyzing those objects' behavior in your reference frame.

The quickest way I can think of to summarize time dilation using that point of view, is: "A clock that is moving with respect to you, runs slower (in your reference frame) than a clock that is stationary with respect to you. The faster it moves, the slower it runs. But in the reference frame of another observer who is moving along with the clock, it always runs at its normal rate."

 

[pervect]

so in english, what is the theory of relativity?

There are several different ways of describing relativity. The simplest treatment that I'm aware of is "Al's relativistic adventure" at http://www.onestick.com/relativity/ (this is the winning entry to Pirelli's relativity challenge, the challenge being to explain relativity as simply as possible).

 

[notknowing]

In plain English, Special Relativity says:

1. The laws of the universe in a non-accelerating reference frame are the same everywhere.
2. The speed of light is constant.

But that doesn't really tell you how time dilation works...

Suppose I consider the famous law E = m c^2.
Is this law violated when observed from an accelerated reference frame ?
(I think so, but I 'd like to have this confirmed)

 

[bernhard.rothenstein]

accelerated E=mcc

Suppose I consider the famous law E = m c^2.
Is this law violated when observed from an accelerated reference frame ?
(I think so, but I 'd like to have this confirmed)

I think not if in
E=m0cc/(1-VV/cc)^1/2 you replace the instantaneous velocity which depends on the accelerated motion you consider (hyperbolic). Probably that answer will generate better answers. We find a simillar situation in the case of the Doppler Effect with accelerating observer and stationary source of light.
We all are here to learn from each other

 

[MeJennifer]

Suppose I consider the famous law E = m c^2.
Is this law violated when observed from an accelerated reference frame ?
(I think so, but I 'd like to have this confirmed)

Actually there is no such thing as an accelerated reference frame.
Furthermore the speed of light for an accelerated observer is no longer c. In the extreme an accelerated observer can even travel in such a way that a light beam following him can never catch up with him.

 

[bernhard.rothenstein]

accelerated reference frame exists?

Actually there is no such thing as an accelerated reference frame.
Furthermore the speed of light for an accelerated observer is no longer c. In the extreme an accelerated observer can even travel in such a way that a light beam following him can never catch up with him.

As far as I know an accelerated reference frame and what takes place inside it is known in special relativity. E.A.Desloge and I.Philpott, "Uniformly accelerated reference frames in special relativity," Am.J.Phys. 55 (3) 252-261 (1987)We can consider the problem in the reference frame considered as stationary and relative to which the body of rest mass m performs the hyperbolic motion its velocity never reaching c and in which light propagates with c.
the best thing a physicist can offer to another one are information and constructive criticism

 

[nakurusil]

Actually there is no such thing as an accelerated reference frame.
Furthermore the speed of light for an accelerated observer is no longer c. In the extreme an accelerated observer can even travel in such a way that a light beam following him can never catch up with him.

Actually you are wrong (Bernhard has already given you another very good reference). See another one http://arxiv.org/PS_cache/physics/pdf/0509/0509161.pdf

 

[MeJennifer]

Actually you are wrong (Bernhard has already given you another very good reference). See another one http://arxiv.org/PS_cache/physics/pdf/0509/0509161.pdf

Really, so for instance an observer who has a proper acceleration of 1G for say 10 seconds can be represented by one single frame? How?

By the way I am not quite sure what the relevance is to the referenced document it seems to deal with the twin paradox.

 

[nakurusil]

Really, so for instance an observer who has a proper acceleration of 1G for say 10 seconds can be represented by one single frame? How?

By the way I am not quite sure what the relevance is to the referenced document it seems to deal with the twin paradox.

A sequence of frames. Try reading the second reference I gave above.

 

[MeJennifer]

A sequence of frames. Try reading the second reference I gave above.

And each single frame is everywhere orthonormal?

 

[notknowing]

Actually there is no such thing as an accelerated reference frame.
Furthermore the speed of light for an accelerated observer is no longer c. In the extreme an accelerated observer can even travel in such a way that a light beam following him can never catch up with him.

I agree that the speed of light for an accelerated observer is no longer c. I'm not sure about the non-existence of an accelerated reference frame. Suppose one replaces "reference frame" by "observer" (since I guess that a frame is used as a tool for an observer), then one can certainly talk of an accelerated observer. Nothing prevents an accelerated observer to make observations of the things around him, just like an observer in constant motion. On the other hand, I'm not sure of the actual rigourous mathematical definition of reference frame, so I could be wrong . Some help of the experts would be useful here.

 

[masudr]

How do you synchronise clocks in an accelerated reference frame?

 

[George Jones]

How do you synchronise clocks in an accelerated reference frame?

The stuff below, which I originally posted on sci.physics, uses the radar method to set up a coordinate system for an accelerated observer. Also, people interested accelerated observers might be interested in https://www.physicsforums.com/showthread.php?t=110742".

Special relativity is robust enough to handle accelerated reference frames. These reference frames share some of the properties of coordinate systems in general relativity, i.e., they're not global, and they often possesses horizons. But they do this within the confines of special relativity.

First, a review of how an inertial observer establishes an inertial coordinate system using her wristwatch and light signals. Suppose [itex]P[/itex] is any event in spacetime. The observer continually sends out light signals, and suppose the light signal that reaches event [itex]P[/itex] left her worldline at time [itex]t_1 [/itex] according to her watch. Upon reception of this signal, [itex]P[/itex] immediately sends a light signal back to the observer, which she receives at time [itex]t_2[/itex]. If [itex]x[/itex] is the spatial distance of [itex]P[/itex] from the observer's worldline, then, since the light goes out and back, the light travels a distance [itex]2x[/itex] in a time [itex]t_2 - t_1[/itex]. Thus

[itex]2x = c (t_2 - t_1)[/itex], or [itex]x = (t_2 - t_1)/2[/itex] with [itex]c = 1[/itex].

The light spends half the time going out, and half the time coming back. Therefore the time coordinate of event [itex]P[/itex] is the same as the the event on the observer's worldline that is halfway (in time) between the observer's emission and reception events. Consequently, the time coordinate of [itex]P[/itex] is

[tex]t = (t_2 + t_1)/2.[/tex]

It is easy to convince oneself that this operational definition establishes a standard inertial coordinate system.

Note that [itex]t_1[/itex] and [itex]t_2[/itex] are proper times for the observer that sets up the coordinate system.

Assume that an accelerated observer uses the same procedure to establish a non-inertial coordinate system. Consider the case where an observer has constant acceleration. Let [itex](t , x)[/itex] be standard coordinates for a global inertial frame. Let the worldline for the accelerated observer be parameterized by her wristwatch (proper) time [itex]T[/itex] so events on her worldline have coordinates

[tex](t , x) = (a^{-1} \mathrm{sinh}(aT) , a^{-1} \mathrm{cosh}(aT))[/tex]

This is one branch of the hyperbola [itex]t^2 - x^2 = - a^{-2}[/itex].

Now set up a coordinate system for the accelerated observer using the light signal procedure given above for an inertial observer. Let [itex]P[/itex] be an event in spacetime that: receives a light signal that left the accelerated observer at proper time [itex]T_1[/itex]; sends a light signal that the accelerated observer receives at time [itex]T_2[/itex]. The accelerated frame coordinates are [itex](t' , x') = ((T_2 + T_1)/2 , (T_2 - T_1)/2)[/itex].

To get a handle on this coordinate system, find what curves of constant [itex]x'[/itex] and curves of constant [itex]t'[/itex] look like in the inertial coordinate system. Let [itex]Q[/itex] and [itex]R[/itex] be the emission and reception events of the accelerated observer, respectively. The inertial coordinates of [itex]P[/itex], [itex]Q[/itex], and [itex]R[/itex] are

[tex]P: (t , x) , Q: ((a^{-1} \mathrm{sinh}(aT_1) , a^{-1} \mathrm{cosh}(aT_1)),[/tex]

[tex]R: (a^{-1} \mathrm{sinh}(aT_2) , a^{-1} \mathrm{cosh}(aT_2))[/tex]

[itex]QP[/itex] lightlike gives

[tex]t - a^{-1} \mathrm{sinh}(aT_1) = x - a^{-1} \mathrm{cosh}(aT_1),[/tex]

which gives

[tex]t - x = (a^{-1}) (\mathrm{sinh}(aT_1) - \mathrm{cosh}(aT1)) = -(a^{-1}) e^(-aT_1).[/tex]

Similarly, [itex]PR[/itex] lightlike gives

[tex]t + x = a^{-1} e^{aT_2}.[/tex]

Multiplying these equations gives

[tex]t^2 - x^2 = -a^{-2} e^{a(T_2 - T_1)} = -a^{-2} e^{2ax'},[/tex]

so curves of constant [itex]x'[/itex] are hyperbolae in the inertial coordinates. The hyperbolae all have asymptotes [itex]t = x[/itex] and [itex]t = -x[/itex]. For [itex]x > 0[/itex], these hyperbolae successively become less sharply curved as [itex]x[/itex] increases.

Dividing the equations gives

[tex]t = \frac{e^{2at'} - 1}{e^{2at'} + 1}x,[/tex]

so curves of constant [itex]t'[/itex] are straight lines that pass through the origin of the inertial coordinates. As [itex]t' \rightarrow \infty[/itex] the lines approach [itex]t = x[/itex]; as [itex]t' \rightarrow -\infty[/itex], the lines approach [itex]t = -x[/itex].

The accelerated coordinate system [itex](t' , x')[/itex] only covers the wedge [itex]x > |t|[/itex] of the global inertial coordinate system, with the halflines [itex]t = x[/itex] and [itex]t = -x[/itex] playing the role of horizons.

These coordinates are related to Rindler coordinates, which when used for normal modes for quantum field theory give rise to the Unruh effect.

I wish I could have included diagrams. Misner, Thorne, and Wheeler, in 6.6 The Local Coordinate System of an Accelerated Observer, use a different method to derive this coordinate system. Their figure 6.4 illustrates the situation nicely.

PS I owe you a post in another thread.

 

[notknowing]

The stuff below, which I originally posted on sci.physics, uses the radar method to set up a coordinate system for an accelerated observer. Also, people interested accelerated observers might be interested in https://www.physicsforums.com/showthread.php?t=110742".

Special relativity is robust enough to handle accelerated reference frames. These reference frames share some of the properties of coordinate systems in general relativity, i.e., they're not global, and they often possesses horizons. But they do this within the confines of special relativity.

Thank you George for clarifying this difficult topic.
Coming back to an earlier question in this thread: It is always mentioned explicitly (in textbooks) that the laws of physics should remain the same for observers moving at CONSTANT speed relative to each other. In some cases, some "laws" such as the Doppler effect or the time dilatation seem to remain the same provided one uses the instantaneous velocity in these expressions.
However, I can not believe that this is generally true (for any law, such as E=mc^2, etc.). Why should one otherwise stress with so many words that one should consider inertial observers only ? Could you comment on this ?

 

[MeJennifer]

Then why do they write in MTW:

At a certain distance from the accelerated world line, successive spacelike hyperplanes, instead of advancing with increasing [itex] \tau[/tex], will be retrogressing. At this distance, and at greater distances, the concept of "coordinates relative to the accelerated observer" becomes ambigious and has to be abandoned.

My understanding is that a reference frame from an acclerated observer is only valid locally in the wedge between the halflines [itex]t = x[/itex] and [itex]t = -x[/itex].

I am wrong?

The accelerated frame coordinates are [itex](t' , x') = ((T_2 + T_1)/2 , (T_2 - T_1)/2)[/itex].

Could you explain how you can justify simply dividing T_2 + T_1 and T_2 - T_1 by two. While I understand that this is valid for an inertial observer, it seems to me that that procedure would give a false result for an accelerated observer. Clearly the speed of light is no longer isotropic for an accelerated observer.

 

[bernhard.rothenstein]

Thank you George for clarifying this difficult topic.
Coming back to an earlier question in this thread: It is always mentioned explicitly (in textbooks) that the laws of physics should remain the same for observers moving at CONSTANT speed relative to each other. In some cases, some "laws" such as the Doppler effect or the time dilatation seem to remain the same provided one uses the instantaneous velocity in these expressions.
However, I can not believe that this is generally true (for any law, such as E=mc^2, etc.). Why should one otherwise stress with so many words that one should consider inertial observers only ? Could you comment on this ?

i comment the case of the Doppler effect with accelerating observer and stationary source. In order to measure the period the accelerated observer should detect two successive light signals (wave crests), detecting them from two different points (non-locality in the period measurement by an accelerating observer). During that time interval the velocity of the observer
and the angle under which he receives the two successive light signals could change. if we make, as many authors do, the "very small period, very high frequency assumption" then we can consider that the observer receives the two successive wave crests remaining located at the same point in space, nothing changes and we obtain an approximate result introducing in the formula that accounts for the Doppler Effect the instantaneous velocity. in the case of the aberration formula, using the instantaneous velocity we obtain correct results because the measurement of the angle can be done continuously. in the paper
W. Cochran, "Some results on the relaivistic Doppler Effect for accelerfated motion," Am.J.Phys. 57, 1039-1043 (19890 the very small period assumption is made whereas in the paper
William Moreau, "Nonlocality in frequency measurements of uniformly accelerating observers," Am.J.Phys. 60, 561-564 (1992) the mentioned nonlocality is taken into account.

 

[Chris Hillman]

In English?

so in english, what is the theory of relativity?

metrology = horology

Oh wait, that was an equation, wasn't it? In faux Graeco-Latin? Never mind.

(Metrology is properly the science of distance measurements, while horology is the science of time measurements. If anyone wants to follow up on this jest, see the book by Peter Galison: Einstein's clocks, Poincare's maps.)

 

[bernhard.rothenstein]

what is relativity?

In plain English, Special Relativity says:

1. The laws of the universe in a non-accelerating reference frame are the same everywhere.
2. The speed of light is constant.

But that doesn't really tell you how time dilation works...

i would add, not in the best English, that special relativity is a theory which prevents us from finding out experiments or formulas enabling us, confined in a laboratory, to distinguish if we are in a state of rest or in a state of uniform motion.
sine ira et studio

 

[DaveC426913]

i would add, not in the best English, that special relativity is a theory which prevents us from finding out experiments or formulas enabling us, confined in a laboratory, to distinguish if we are in a state of rest or in a state of uniform motion.
sine ira et studio

That sounds a bit misleading, because it suggests that we are being stopped from discovering this.

It would be better to say that SR demonstrates to us that there is no such thing as a difference between a state of rest and a state of uniform uniform motion.

 

[pervect]

Then why do they write in MTW:

My understanding is that a reference frame from an acclerated observer is only valid locally in the wedge between the halflines [itex]t = x[/itex] and [itex]t = -x[/itex].

I am wrong?Could you explain how you can justify simply dividing T_2 + T_1 and T_2 - T_1 by two. While I understand that this is valid for an inertial observer, it seems to me that that procedure would give a false result for an accelerated observer. Clearly the speed of light is no longer isotropic for an accelerated observer.

A coordinate system really has to do only one thing - that is assign a unique set of coordinates to every given point in some region of space-time.

"Radar coordinates" accomplish this. They do not accomplish some other things you may have come to expect in a coordinate system though. Specifically, if you look at the distances between "tic marcs" spaced at uniform small coordinate intervals along the x axis, i.e for instance (t=0, x=0), (t=0,x=.1), (t=0, x=.2) etc, you will find that the spacing of these tic marcs ins't uniform.

(How do you find the spacing between tic marks? If they are close enough, you just find the Lorentz interval between them. If they are not close enough, you have to integrate over a curve of constant time.)

However, the non-uniformity of the tic marks is not a fatal flaw.

I believe that your arguments about the Rindler wedge (between t=x and t=-x) are correct as stated for Rindler coordinates, but not for radar coordinates. Note that radar coordinates will simply not assign numbers to the region of space-time that is not causally connected. So the radar coordinate system is still local - only some points get coordinates in "radar coordinates", the points that are not causally connected do not get assigned coordinates.

 

[bernhard.rothenstein]

That sounds a bit misleading, because it suggests that we are being stopped from discovering this.

It would be better to say that SR demonstrates to us that there is no such thing as a difference between a state of rest and a state of uniform uniform motion.

thanks. i mentioned that what I say is not in the best English. Galileo inviting us to be confined in a laboratory helped us to discover this. I always started to teach SR with mentioning Galileo's description of what hapens in an inertial reference frame at rest or in uniform motion.

 

[EmilK]

Caltech provides an interesting video introduction on General Relativity the can be found here.