Inverse Funcation of Relativistic mass

[Uncle6]

The relativistic mass equation shows the relationship between relativistic mass and speed. Basically showing that the closer u get to light speed the less relativistic mass you have, but ur rest mass will be the same.

I was asked by my friend what the inverse of that formula means.

Thus I took the original equation and came up with the inverse: m = sqrt(c^2-(m0c^2/v)^2)

I also looked at the graph of the original and the inverse would just be a reflection about y=x.

I still can't interpret it. Any hints/ help?

 

[jtbell]

The relativistic mass equation shows the relationship between relativistic mass and speed.

Do you mean

[tex]m = \frac{m_0}{\sqrt{1 - v^2 / c^2}}[/tex]

Basically showing that the closer u get to light speed the less relativistic mass you have, but ur rest mass will be the same.

Don't you mean "more relativistic mass" (not "less")?

Thus I took the original equation and came up with the inverse: m = sqrt(c^2-(m0c^2/v)^2)

You must have made a mistake in your algebra somewhere. The units of the various quantities don't work out consistently in that equation. Also, if you started with the equation I gave above, an "inverse" equation would not start with "m =" but with "v =".

 

[Uncle6]

I switched m for v and isolated for m

 

[Fredrik]

We were supposed to guess that the relativistic mass in your equation was actually a velocity!? It's always a bad idea to change the meaning of the symbols without telling anyone.

If you solve the equation in jtbell's post for v, you get

[tex]v=c\sqrt{1-\frac{m_0^2}{m^2}}=c\sqrt{1-\frac{m_0^2c^4}{E^2}}[/tex]

where E=mc² is the energy. Note that v→c when E→∞.