# Homework Help: Special Relativity - Relativistic Dynamics

1. Apr 16, 2010

### malindenmoyer

I am trying to understand a section in a textbook I have regarding Special Relativity, specifically deriving an expression for what is known as "relativistic mass", in order to find an expression for "relativistic momentum". I have attached the pages in the book which are giving me trouble, more specifically, the paragraph I have highlighted in red. I understand everything before that, however I can't seem to understand what the paragraph is saying.

I have derived an expression for the Lorentz Velocity Transformation so I understand where the author is getting at, just don't exactly recognize how the numerator of the transformation equation is 2u assuming my derived expression is:

$$v_x=\frac{v_x^{'}+u}{1+\frac{u v_x^'}{c^2}}$$

Can somebody restate what he is saying in the highlighted paragraph?

[PLAIN]http://people.tamu.edu/~malindenmoyer/tamu/special_relativity.png [Broken]

Reference:

Taylor, J. G. Special Relativity. Oxford: Clarendon, 1975. Print.

Last edited by a moderator: May 4, 2017
2. Apr 17, 2010

### thebigstar25

well, what i understand is that the expression for the Lorentz Velocity Transformation you provided gives the relationship between the velocities of a particle measured by two observers .. one of them in a O (just an example) frame, where he sees the particle moving with velocity vx, and the other one in the O frame, where he sees the particle moving with vx ..

so going back to your example:
lets say we are considering the particle on the left side ..

for an observer in the S frame (which equivalent to an observer in the O frame), this particle has velocity of U
and for an observer in the S frame (which equivalent to an observer in the O fram), it has a velocity of u

so using that expression for the Lorentz Velocity Transformation, you can see that vx is substituted with U and vx is substituted with u, yielding to the equation in the text..

3. Apr 17, 2010

### malindenmoyer

My only question now is that S' is moving to the left relative to S, and the particle on the left is moving to the right, both with velocity u. However, this implies that the numerator would essentially cancel out, as the signs are opposite. What am I missing?

4. Apr 17, 2010

### thebigstar25

i got your point now ..

(im not sure from my answer) .. but the only why to get the result they got is by considering this equation:

vx = vx - u / (1 - vx.u/c^2), with vx` = U, vx = u, and u = -u ..

I hope im not confusing you, but that what i can think of for now ..

5. Apr 17, 2010

### malindenmoyer

I think that is the solution...it's a little confusing what to refer as vx' and vx....but I think I understand it now. Thanks!

6. Apr 17, 2010

### thebigstar25

i wished i could give a convincing explanation .. I hope other member would give better explanation for your question ..