# Problem understanding relativistic momentum

Many books teaching relativity at the introductory level,go on and say that relativistic momentum cant be described using p=mv

and p=m Dx/Dt(D is Delta)

also

it says the measure Dx is the same as measured by the both people.One who is measuring his own momentum and another who is measuring the moving person's(with respect to him) momentum.

I get confused.how can this be?

Hootenanny
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Gold Member
Many books teaching relativity at the introductory level,go on and say that relativistic momentum cant be described using p=mv

and p=m Dx/Dt(D is Delta)

also

it says the measure Dx is the same as measured by the both people.One who is measuring his own momentum and another who is measuring the moving person's(with respect to him) momentum.

I get confused.how can this be?
Perhaps it is meant to say something like the following;

"The above equations assume that the measure[ment] Dx is the same as measured by the both people.One who is measuring his own momentum and another who is measuring the moving person's(with respect to him) momentum."

Why shd they both be the same?As the person will refer to another thing to measure Dx.Like the length of road he covers.But the road will seem contracted to him isnt it?

May be quoting the text would help clear things up.

Hootenanny
Staff Emeritus
Gold Member
Why shd they both be the same?As the person will refer to another thing to measure Dx.Like the length of road he covers.But the road will seem contracted to him isnt it?
They are not the same in special relativity, the quantity dx will indeed be subject to length contraction as described by the lorentz transformations. However, the two equations you have quoted are the equations for classical momentum which are based on the Galilean transformations and hence dx is not subject to length contraction.

Heres the text quoted:
Consider a particle moving with a constant speed v in the positive x direction.Classically it has momentum=mv=mDx/Dt

in which Dx is the distance it travels in time Dt.To find a relativistic expression for momentum,we start with the new definition

p=mDx/Dt(o)

Here,as before Dx is the distance travelled by a moving particle as viewed by the observer watching that particle.However t(o) is the time required to travel that distance,measured by not the observer watching the moving particle but by an observer moving with the particle.

Using time dilation t/gamma=t(o)

therefore p=mv[gamma]

I didnt understand why distance Dx is same for both observers!

Doc Al
Mentor
I didnt understand why distance Dx is same for both observers!
Where in that quoted passage does it say that Dx is the same for both observers? It says: "Dx is the distance travelled by a moving particle as viewed by the observer watching that particle". Nowhere does it say that different observers measure the same distance.

How is the textbook then comparing the two momentum(S) measured by the person moving with the particle and the person who is not at rest with respect to the particle?

Hootenanny
Staff Emeritus
Gold Member
How is the textbook then comparing the two momentum(S) measured by the person moving with the particle and the person who is not at rest with respect to the particle?
Time dilation, note the term t0, the proper time interval.

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The argument as per the book:

Rest Frame:

$$p_{0} = m\frac{\Delta x}{\Delta t_0}$$

Moving frame:

$$\Delta t = \gamma\Delta t_0$$

$$\therfore p_{0} = \gamma m\frac{\Delta x}{\Delta t}$$

But shouldn't it be...

Rest Frame:

$$p_{0} = m\frac{\Delta x_0}{\Delta t_0}$$

Moving frame:

$$\Delta t = \gamma\Delta t_0$$

$$\Delta x = \gamma^{-1}\Delta x_0$$, and therfore

$$p = m\frac{\Delta x}{\gamma^2\Delta t}$$
?

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Hootenanny
Staff Emeritus
Gold Member
But shouldn't it be...
Rest Frame:
$$p_{0} = m\frac{\Delta x_0}{{\color{red}\Delta t_0}}$$
Are you sure about that? Read again, t0 is not in the rest frame of the observer, but in that of the particle.

Oh, okay. So from the particle's frame only length contraction is observed. So that would lead to only one gamma factor.

im confused
Can som1 please explain well.I am asorry ive tried a lot but cudnt understand.Why would $$\Delta x$$ be the same for both?

Hootenanny
Staff Emeritus
Gold Member
Oh, okay. So from the particle's frame only length contraction is observed. So that would lead to only one gamma factor.
Correct. Indeed, one can do equivalent procedures by considering length contraction or time dilation. The particle experiences length contraction, whereas the observer experiences time dilation; both will lead to the same results, provided the principles are applied correctly.
im confused
Can som1 please explain well.I am asorry ive tried a lot but cudnt understand.Why would $$\Delta x$$ be the same for both?
No one has ever said they are! $\Delta x$ is the distance measured by the observer not by the particle.

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Explanation please.I still cant get it :'(

Thanks, Hoot.

Hootenanny
Staff Emeritus
Gold Member
Explanation please.I still cant get it :'(
What exactly don't you understand?
Thanks, Hoot.
No problem

Why length contraction isnt observed!Also which part of neutrino's equations were correct?Thnx

Hootenanny
Staff Emeritus
Gold Member
Why length contraction isnt observed!Also which part of neutrino's equations were correct?Thnx
Length contraction is observed, just not by an observer.
Correct. Indeed, one can do equivalent procedures by considering length contraction or time dilation. The particle experiences length contraction, whereas the observer experiences time dilation; both will lead to the same results, provided the principles are applied correctly.
Also which part of neutrino's equations were correct?Thnx
The first part under the heading 'According to the Book' is correct.

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O boy i just cant seem to get it

According to me:

We are comparing the momentum an observer measures of a moving body and which the moving body measures of itself.Now suppose the moving body is on a road and has two stones which have a distance of Dx(with respect to the observing frame,i mean the rest observer).

Now when the observer measures the body's momentum:
$$p = m\frac{\Delta x}{\Delta t}$$ ..........(1)

also when the moving body measures his own momentum
$$p_{0} = m\frac{\Delta x_0}{\Delta t_0}$$
but as
$$\Delta x_0 = \gamma^{-1}\Delta x$$

Now im getting:
$$p_{0} = m\frac{\Delta x}{\gamma^2\Delta t}$$

So thats where im going wrong!:'((((((((((((((

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Hootenanny
Staff Emeritus
Gold Member
Now when the observer measures the body's momentum:
$$p = m\frac{{\color{red}\Delta x}}{\Delta t}$$ ..........(1)
This length is a proper length, since it is measured in the rest frame of your two stones.
also when the moving body measures his own momentum
$$p_{0} = m\frac{{\color{red}\Delta x_0}}{\Delta t_0}$$
The length measure by the moving body is not a proper length. In your example above, the particle is moving relative to the distance between the two stones, therefore the distance between the two stones appears contracted!

err i think i said that$$\Delta x_0 = \gamma^{-1}\Delta x$$

and $$p_{0}$$ is the momentum as measured by the moving body

Hootenanny
Staff Emeritus
Gold Member
err i think i said that$$\Delta x_0 = \gamma^{-1}\Delta x$$

and $$p_{0}$$ is the momentum as measured by the moving body
You did indeed say that. However, the two statements I highlighted are still incorrect for the reasons I stated above.

err.... how?

Hootenanny
Staff Emeritus