Derivation of Velocity-Addition Formula

In summary: There is absolutely no "messy" algebra in what I suggested. The velocity addition formula for the x-direction follows directly and immediately when you do as I suggested. You can also get the velocity addition formulas for the y- and z-directions directly by dividing the relationships in these directions by the relationship for the time direction. You can then substitute these results into the equation for the time direction to confirm that the solution satisfies the time equation.
  • #1
Trifis
167
1
There was an old thread but for some reason it is now closed...

So I'd have to restate this question:

Why does any textbook doesn't bother to derive the 3-vector-velocities addition formula using the more general 4-vector formalism?

In detail, the Lorentz Transformation: U[itex]^{1'}[/itex]=γ(v)(U[itex]^{1}[/itex]-β(v)U[itex]^{0}[/itex])⇔γ(υ[itex]^{'}_{x}[/itex])υ[itex]^{'}_{x}[/itex]=γ(v)[γ(υ[itex]_{x}[/itex])(υ[itex]_{x}[/itex]-cβ(v))] (where υ[itex]_{x}[/itex] the velocity in Σ,υ[itex]^{'}_{x}[/itex] the velocity mesured in Σ', and v the relative velocity between the two systems) solved for υ[itex]^{'}_{x}[/itex] should grant υ[itex]^{'}_{x}[/itex]=(υ[itex]_{x}[/itex]-v)/(1-vυ[itex]_{x}[/itex]/c[itex]^{2}[/itex]) right?
 
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  • #2
Trifis said:
There was an old thread but for some reason it is now closed...

So I'd have to restate this question:

Why does any textbook doesn't bother to derive the 3-vector-velocities addition formula using the more general 4-vector formalism?

In detail, the Lorentz Transformation: U[itex]^{1'}[/itex]=γ(v)(U[itex]^{1}[/itex]-β(v)U[itex]^{0}[/itex])⇔γ(υ[itex]^{'}_{x}[/itex])υ[itex]^{'}_{x}[/itex]=γ(v)[γ(υ[itex]_{x}[/itex])(υ[itex]_{x}[/itex]-cβ(v))] (where υ[itex]_{x}[/itex] the velocity in Σ,υ[itex]^{'}_{x}[/itex] the velocity mesured in Σ', and v the relative velocity between the two systems) solved for υ[itex]^{'}_{x}[/itex] should grant υ[itex]^{'}_{x}[/itex]=(υ[itex]_{x}[/itex]-v)/(1-vυ[itex]_{x}[/itex]/c[itex]^{2}[/itex]) right?

You're exactly right: If you know about the 4-velocity (or equivalently, the Energy-Momentum 4-vector), then you can derive the transformation law for velocities very easily.

However, you can also do it the other way around: You derive the transformation law for velocities using the Lorentz transformations, and then you derive the relativistic form for momentum from that.
 
  • #3
stevendaryl said:
However, you can also do it the other way around: You derive the transformation law for velocities using the Lorentz transformations, and then you derive the relativistic form for momentum from that.

I wouldn't say this is correct, as you'd have to make assumptions about relativistic momentum in order to do this.
 
  • #4
jcsd said:
I wouldn't say this is correct, as you'd have to make assumptions about relativistic momentum in order to do this.

Sure, you need to make assumptions, such as reducing to the Newtonian case as v/c --> 0, and being proportional to mass, and being in the same direction as the velocity, and so forth.
 
  • #5
At first glance the derivation seems straightforward but when put into practice I have some issues with the algebra:

[itex]\frac{u'}{\sqrt{1-(u'/c)^{2}}}[/itex]=[itex]\frac{1}{\sqrt{1-(v/c)^{2}}}[/itex][itex]\frac{1}{\sqrt{1-(u/c)^{2}}}[/itex](u-[itex]\frac{v}{c}[/itex])

The above solved for u' would never yield the velocity-addition formula:frown:

What am I missing?
 
  • #6
Trifis said:
At first glance the derivation seems straightforward but when put into practice I have some issues with the algebra:

[itex]\frac{u'}{\sqrt{1-(u'/c)^{2}}}[/itex]=[itex]\frac{1}{\sqrt{1-(v/c)^{2}}}[/itex][itex]\frac{1}{\sqrt{1-(u/c)^{2}}}[/itex](u-[itex]\frac{v}{c}[/itex])

The above solved for u' would never yield the velocity-addition formula:frown:

What am I missing?

Your last expression on the right should be simply (u-v). Then the algebra should (laboriously) work out - the rest of your derivation looks correct to me.

[Edit: just worked it through - with the one correction, all the algebra does work out. This is your only mistake.]
 
  • #7
PAllen said:
Your last expression on the right should be simply (u-v). Then the algebra should (laboriously) work out - the rest of your derivation looks correct to me.

[Edit: just worked it through - with the one correction, all the algebra does work out. This is your only mistake.]

PAllen is correct. Now, write down the corresponding relationship for the time direction, and divide your x-direction equation by your t-direction equation. What do you get then?

Chet
 
  • #8
Chestermiller said:
PAllen is correct. Now, write down the corresponding relationship for the time direction, and divide your x-direction equation by your t-direction equation. What do you get then?

Chet

Actually, this isn't necessary. Make the one fix to the equation in post #5, and the velocity addition formula follows by pure (messy) algebra. Easiest is just to plug in:

u' = (u-v)/(1-uv/c^2)

into the equation and see that it is a solution. To be complete, you then need to provide arguments that it is the only solution.
 
  • #9
Thanks!

I see it now. cβ(v) is of course v and not v/c...

Working SR in 3D problems has always had some troublesome algebra though...
 
  • #10
PAllen said:
Actually, this isn't necessary. Make the one fix to the equation in post #5, and the velocity addition formula follows by pure (messy) algebra. Easiest is just to plug in:

u' = (u-v)/(1-uv/c^2)

into the equation and see that it is a solution. To be complete, you then need to provide arguments that it is the only solution.

There is absolutely no "messy" algebra in what I suggested. The velocity addition formula for the x-direction follows directly and immediately when you do as I suggested. You can also get the velocity addition formulas for the y- and z-directions directly by dividing the relationships in these directions by the relationship for the time direction. You can then substitute these results into the equation for the time direction to confirm that the solution satisfies the time equation identically.

Chet
 
  • #11
Chestermiller said:
There is absolutely no "messy" algebra in what I suggested. The velocity addition formula for the x-direction follows directly and immediately when you do as I suggested. You can also get the velocity addition formulas for the y- and z-directions directly by dividing the relationships in these directions by the relationship for the time direction. You can then substitute these results into the equation for the time direction to confirm that the solution satisfies the time equation identically.

Chet

True. However, my point was the equation in #5 (with the one correction) was informationally complete as it stands, and that seemed to be the direction the OP wanted to take. The tradeoff for working with only the x component relationship is more algebra to arrive at velocity addition formula.
 

1. How is the velocity-addition formula derived?

The velocity-addition formula is derived using the principles of special relativity, specifically the laws of motion and the constancy of the speed of light. It involves manipulating equations that describe the relationship between time, distance, and velocity, and accounting for the effects of time dilation and length contraction.

2. What is the significance of the velocity-addition formula in physics?

The velocity-addition formula is significant because it allows us to calculate the velocity of an object from different reference frames. This is important in understanding the effects of relativity, as different observers may measure different velocities for the same object due to their relative motion.

3. Can the velocity-addition formula be applied to all types of motion?

Yes, the velocity-addition formula can be applied to all types of motion, including linear, circular, and accelerated motion. It is a fundamental concept in special relativity and is used in many areas of physics, including astrophysics and particle physics.

4. How does the velocity-addition formula differ from the classical formula for adding velocities?

The classical formula for adding velocities assumes that the speed of light is infinite and does not account for the effects of relativity. The velocity-addition formula takes into account the constant speed of light and the effects of time dilation and length contraction, resulting in different values for added velocities at high speeds.

5. Are there any real-world applications of the velocity-addition formula?

Yes, the velocity-addition formula has many practical applications in modern technology, such as in GPS systems, particle accelerators, and spacecraft navigation. It is also used in theoretical physics to understand the behavior of objects at high speeds and in extreme conditions.

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