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gulsen

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- Thread starter gulsen
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In summary, there is a falling relativistic particle with a known momentum p0 at time t0 and p1 at time t1, measured by an observer on the ground. The only force affecting the particle is gravity and it is assumed to follow a Newtonian model. The relativistic change in momentum can be calculated using the differential equation \frac{d}{dt} \left(\frac{ v(t)}{\sqrt{1-\frac{v^2(t)}{c^2}}}\right)=g and solving for the velocity at time t.

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gulsen

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Andrew Mason

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I am assuming this is some kind of cosmic ray particle with relativistic energy.gulsen said:

I am a little beat tonight, but why would it not just be [itex]\Delta p = p_1 - p_0[/itex]? You measure momentum in the same (earth) frame of reference both times:

[tex]p_0 = \gamma_0m_0v_0[/tex] and later:

[tex]p_1 = \gamma_1m_0v_1[/tex]

AM

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gulsen

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It's a relativistic particle in a Newtonian gravitational potential.It falls freely,so you can say that the accleration is constant and equal to "g"...Solve it or search for uniformly accelerated particle in SR...

Daniel.

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Data

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Yes, you can still use the Newtonian model for gravity. Just set up a differential equation

[tex]\frac{GM}{r^2} = \frac{d^2r}{dt^2}[/tex]

with initial condition

[tex]\frac{dr}{dt}\biggr |_{t=0} = \frac{p_0}{m_0}[/tex]

and solve for [itex]dr/dt[/itex], ie. the velocity, then use your solution to find the velocity at time [itex]t[/itex].

If you are allowed to make the approximation

[tex]\frac{d^2r}{dt^2} = g[/tex]

where [itex]g[/itex] is the acceleration at sea level, then the DE is much simpler.

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No,no,it's a relativistic particle.It has 4 momentum...

[tex] \frac{d}{dt}\left(\frac{1}{\sqrt{1-\frac{v^{2}(t)}{c^{2}}}} v(t)\right) = g [/tex]

Daniel.

[tex] \frac{d}{dt}\left(\frac{1}{\sqrt{1-\frac{v^{2}(t)}{c^{2}}}} v(t)\right) = g [/tex]

Daniel.

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Data

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[tex]\frac{d}{dt} \left(\frac{ v(t)}{\sqrt{1-\frac{v^2(t)}{c^2}}}\right)=g[/tex]

but other than that, you're actually probably right. It might expect you to use that form for the DE. Time to dig up SRT books~

The concept of relativistic change in momentum is based on Einstein's theory of relativity, which states that the laws of physics are the same for all observers in uniform motion. It is the change in momentum of an object moving at speeds close to the speed of light, taking into account the effects of time dilation and length contraction.

Relativistic change in momentum differs from classical change in momentum in that it takes into account the effects of special relativity, such as time dilation and length contraction. In classical mechanics, momentum is simply mass times velocity, but in relativity, the mass of an object is not constant and its velocity cannot exceed the speed of light.

The equation for calculating relativistic change in momentum is given by **p = mv / √(1 - v^2/c^2)**, where p is the relativistic momentum, m is the mass of the object, v is its velocity, and c is the speed of light.

No, relativistic change in momentum is only significant at speeds close to the speed of light, which is much higher than the speeds we encounter in everyday life. It is mainly observed in extreme environments, such as particle accelerators or astronomical events involving high speeds.

Relativistic change in momentum is important in the field of particle physics, where it is used to understand the behavior of subatomic particles at high speeds. It also plays a crucial role in technologies such as GPS, which rely on the precise measurement of time and space, taking into account relativistic effects.

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