# V=? for Relativistic Mass,length contraction & time dilation

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1. Dec 6, 2016

### Axidecimal

1. The problem statement, all variables and given/known data
Velocity Equations for Relativistic Mass,length contraction and time dilation.
I was able to figure out one. This is not for homework. I want to learn these equations for future reference.

2. Relevant equations

3. The attempt at a solution

Length Contraction : v = c √{1-(l'/lo)^2}
Time Dilation: ?
Mass: ?

2. Dec 6, 2016

### Simon Bridge

Just so I understand how you are thinking so I can answer properly:
How did you work out the length-contraction equation? Why not do the same thing for time dilation?

Note: no such thing as "relativistic mass". Used to be a thing but it turns out not to be much of a useful concept.

3. Dec 6, 2016

### Axidecimal

l'/lo = √{1-v^2/c^2}
(l'/lo)^2 = 1-v^2/c^2
1-(l'/lo)^2 = v^2/c^2
√{1-(l'/lo)^2} = v/c
c√{1-(l'/lo)^2} = v

Mass increase has not been discontinued in my program.

Last edited: Dec 6, 2016
4. Dec 7, 2016

### Simon Bridge

Why not do the same thing for time dilation?

OK you need to do relativistic mass for passing exams - just understand that it is an out of date concept.

5. Dec 7, 2016

### Axidecimal

I just copied the equations from wiki to this platform but i now have come to realize they are the same.
This is pretty self explanatory. i was given equations from my lessons that were essentially the same but all look different so i got confused. I didnt even look at the wiki equations closely until now

Last edited: Dec 7, 2016
6. Dec 7, 2016

### Orodruin

Staff Emeritus
Better still: Point your teachers to my Insight post What is relativistic mass and why it is not used much? or refer them to me. It is simply a concept that is not used in physics today and it is really just confusing people to use it. Unfortunately, the concept permeates much of the introductory physics literature - which often is not written by people with specialist knowledge on relativity - and therefore becomes used by teachers at pre-university and introductory university level.

7. Dec 7, 2016

### Simon Bridge

Well done.
The form of the equations is $x = \frac{y}{\sqrt{1-z^2}}$ and you want to solve for $z$.
It does not matter what the actual letters are.

It is more important to understand how to use the equations - that is the hard part to get your head around.