How is the Relativity Formula Proven for an Electron in an Electric Field?

[Samkiwi]

Homework Statement

I'm having trouble finding the proof of the relativistic acceleration formula starting from the velocity formula, I've been working on it for a long time but I can't solve this question. :)

Relevant Equations

electromagnetism and relativity

It is an electron initially pushed by the action of the electric field. The vectors of force and velocity are parallel to each other.

Here's the questionA possible expression of speed as a function of time is the following:

$$v(t) = \frac{At}{\sqrt{1 + (\frac{At}{c})^2}}$$where is it $$A =\frac{qE}{m}$$
Taking into account that [2] can be written in the equivalent form.
$$\frac{dv}{dt}=\frac{qE}{m}(1-\frac{v^{2}}{c^{2}})^{-\frac{3}{2}}[3]$$
verify by deriving and substituting that the function v (t) defined by [2] satisfies [3]

 

[PeroK]

What do you get when you differentiate $v(t)$?

 

[Samkiwi]

i get this

$$\frac{dv}{dt}=(1+(\frac{At}{c})^{2})^{-\frac{3}{2}}(\frac{At}{c})^2$$

 

[PeroK]

i get this

$$\frac{dv}{dt}=(1+(\frac{At}{c})^{2})^{-\frac{3}{2}}(\frac{At}{c})^2$$

You need to use the product rule. I'm not sure how you got that. As a first step you should have something like:
$$\frac{dv}{dt} = \frac{A}{\sqrt{1 + (\frac{At}{c})^2}} + \dots$$

 

[Samkiwi]

and i know that

$$A=\frac{qE}{m}$$
and
$$At=\frac{qE}{m}t$$

$$F=maϒ^3$$

then

$$A=aϒ^3$$

 

[Samkiwi]

yes yes I was wrong I considered t a constant, when in reality it is a variable

You need to use the product rule. I'm not sure how you got that. As a first step you should have something like:
$$\frac{dv}{dt} = \frac{A}{\sqrt{1 + (\frac{At}{c})^2}} + \dots$$

yes yes I was wrong I considered t a constant, when in reality it is a variable

 

[Samkiwi]

You need to use the product rule. I'm not sure how you got that. As a first step you should have something like:
$$\frac{dv}{dt} = \frac{A}{\sqrt{1 + (\frac{At}{c})^2}} + \dots$$

I obtain this

$$\frac{dv}{dt}=\frac{Ac^{3}}{(c^{2}+A^{2}t^{2})\sqrt{c^{2}+A^{2}t^{2}}}$$

and now?

 

[PeroK]

I obtain this

$$\frac{dv}{dt}=\frac{Ac^{3}}{(c^{2}+A^{2}t^{2})\sqrt{c^{2}+A^{2}t^{2}}}$$

and now?

That's correct, although you've changed the format of the term $1 + (\frac{At}{c})^2$ to $c^2 + (At)^2$.

You can simplify what you have to the correct answer. But, it's not good technique to change things that start in the format required for the final answer to something else that looks different.

 

[Samkiwi]

okay, now I try, anyway thank you very much yes it was very useful I don't know how to thank you

That's correct, although you've changed the format of the term $1 + (\frac{At}{c})^2$ to $c^2 + (At)^2$.

You can simplify what you have to the correct answer. But, it's not good technique to change things that start in the format required for the final answer to something else that looks different.

 

[PeroK]

okay, now I try, anyway thank you very much yes it was very useful I don't know how to thank you

I see now there is a further step needed. What you should get, when you simplify that expresson is:
$$\frac{dv}{dt} = A(1 + (\frac{At}{c})^2)^{-\frac 3 2 }$$
And, what you are aiming for is: $$\frac{dv}{dt} = A(1 - (\frac{v}{c})^2)^{\frac 3 2 }$$
That means that you to show: $$\frac{1}{1 + (\frac{At}{c})^2} = 1 - (\frac{v}{c})^2$$

 

[Samkiwi]

I see now there is a further step needed. What you should get, when you simplify that expresson is:
$$\frac{dv}{dt} = A(1 + (\frac{At}{c})^2)^{-\frac 3 2 }$$
And, what you are aiming for is: $$\frac{dv}{dt} = A(1 - (\frac{v}{c})^2)^{\frac 3 2 }$$
That means that you to show: $$\frac{1}{1 + (\frac{At}{c})^2} = 1 - (\frac{v}{c})^2$$

I didn't understand how I should demonstrate the last step

 

[PeroK]

I didn't understand how I should demonstrate the last step

It's a relatrively simple piece of algebra. Just calculate, using the expression you have for $v$, the quantity: $$1 - \frac{v^2}{c^2}$$

 

[Samkiwi]

It's a relatrively simple piece of algebra. Just calculate, using the expression you have for $v$, the quantity: $$1 - \frac{v^2}{c^2}$$

I may sound silly, but would you explain each step please, sorry I'm nagging

 

[PeroK]

I may sound silly, but would you explain each step please, sorry I'm nagging

You have to make an effort yourself. The first step is easy, so I'll do that: $$v^2 = \frac{(At)^2}{1 + (\frac{At}{c})^2}$$

 

[Samkiwi]

You have to make an effort yourself. The first step is easy, so I'll do that: $$v^2 = \frac{(At)^2}{1 + (\frac{At}{c})^2}$$

and after?

 

[PeroK]

and after?

That's up to you.

 

[Samkiwi]

That's up to you.

I'm not understanding anything anymore

differentiating I found this

$$\frac{dv}{dt}=A\frac{c^{6}}{(\sqrt{c^{2}+v^{2}})^{3}}$$

or this

$$\frac{dv}{dt}=Ac^{3}(1+\frac{v^{2}}{c^{2}})^{-\frac{3}{2}}$$

 

[PeroK]

If $$v^2 = \frac{(At)^2}{1 + (\frac{At}{c})^2}$$ then what is $$1 - \frac{v^2}{c^2}$$?

 

[Samkiwi]

If $$v^2 = \frac{(At)^2}{1 + (\frac{At}{c})^2}$$ then what is $$1 - \frac{v^2}{c^2}$$?

$$\frac{c^{2}}{c^{2}+A^{2}t^{2}}$$

and now?

 

[PeroK]

That's it, isn't it? Apart from changing the format of the expresson.

 

[Samkiwi]

That's it, isn't it? Apart from changing the format of the expresson.

no, the form of the expression asking the question is this
$$\frac{dv}{dt}=\frac{qE}{m}(1-\frac{v^{2}}{c^{2}})^{-\frac{3}{2}}$$

 

[PeroK]

no, the form of the expression asking the question is this
$$\frac{dv}{dt}=\frac{qE}{m}(1-\frac{v^{2}}{c^{2}})^{-\frac{3}{2}}$$

See post #10. Note that $A = \frac{qE}{m}$.

 

[PeroK]

no, the form of the expression asking the question is this
$$\frac{dv}{dt}=\frac{qE}{m}(1-\frac{v^{2}}{c^{2}})^{-\frac{3}{2}}$$

You should check this, as it should be $$\frac{dv}{dt}=\frac{qE}{m}(1-\frac{v^{2}}{c^{2}})^{\frac{3}{2}}$$
PS you can see that it must be this, as we need $\frac{dv}{dt} \rightarrow 0$ as $v \rightarrow c$.

 

[Samkiwi]

$$You should check this, as it should be $$\frac{dv}{dt}=\frac{qE}{m}(1-\frac{v^{2}}{c^{2}})^{\frac{3}{2}}$$
PS you can see that it must be this, as we need $\frac{dv}{dt} \rightarrow 0$ as $v \rightarrow c$.

I was able to find this formula

$$\frac{dv}{dt}=A(1+\frac{v^{2}}{c^{2}})^{-\frac{3}{2}}$$

the only problem is this part

$$(1+\frac{v^{2}}{c^{2}})^{-\frac{3}{2}}$$

I tried what you said. but I don't solve anything like that.
i should make sure that.

$$(1+\frac{v^{2}}{c^{2}})^{-\frac{3}{2}}=(1-\frac{v^{2}}{c^{2}})^{\frac{3}{2}}$$

but how do I prove it?
if I substitute v for the initial formula, I get

$$\frac{\sqrt{(c^{2}+v^{2})^{3}}}{\sqrt{(c^{2}+2v^{2})^{3}}}$$

 

[PeroK]

I was able to find this formula

$$\frac{dv}{dt}=A(1+\frac{v^{2}}{c^{2}})^{-\frac{3}{2}}$$

That can't be right. Just take the limit as $v \rightarrow c$.

Instead, we got as far as this:

If $$v^2 = \frac{(At)^2}{1 + (\frac{At}{c})^2}$$ then what is $$1 - \frac{v^2}{c^2}$$?

You need to pick up from there.

 

[Samkiwi]

That can't be right. Just take the limit as $v \rightarrow c$.

Instead, we got as far as this:You need to pick up from there.

ok he succeeded, thanks a lot