Accelerations in Special Relativity

A

Anamitra Palit

Registered Senior Member
We start with the fact that the norm-square of the four acceleration vector is negative or zero.
Indeed
\[c^2=c^2\left(\frac {dt}{d \tau}\right)^2-\left(\frac {dx}{d \tau}\right)^2-\left(\frac {dy}{d \tau}\right)^2-\left(\frac {dz}{d \tau}\right)^2\] (1)
Differentiating both sides with respect to time we obtain
\[c^2\frac{d^2 t}{dt^2}\frac{dt}{d\tau}-\frac{d^2 x}{dt^2}\frac{dx}{d\tau}-\frac{d^2 y}{dt^2}\frac{dy}{d\tau}-\frac{d^2 z}{dt^2}\frac{dz}{d\tau}=0\] (2)
We transform to an inertial frame where the particle is momentarily at rest.
\[c^2\frac{d^2 t}{d\tau^2}\frac{dt}{d \tau}=0\]
\[\Rightarrow \frac{d^2t}{d \tau^2}=0\](3)
[the above holds since v.v=c^2]
Norm square of the acceleration vector in the new frames of reference [that is on transformation ] is negative or zero. Therefore due to the conservation of dot product it is zero or negative in all (inertial) frames of reference: ||a||<=0
[That the norm square of the acceleration four vector is negative may ve provd by alternative techniques also]
We now consider an accelerating particle executing a one dimensional motion in the x direction
\[v_p=v_x=\frac{dx}{dt}\](4)
Proper acceleration component
\[a_t=\frac{d^2 t}{d \tau^2}=\frac{d}{d\tau}\left(\frac{dt}{d\tau}\right)=\frac {d \gamma_p}{d \tau}\\=\gamma_p^3 \frac{v_p}{c^2}\frac{d v_p}{d \tau}= \gamma_p^3 \frac{v_p}{c^2}\frac{d }{d \tau}\frac{dx}{dt }\]
(5)
Now,
\[\frac{d }{d \tau}\frac{dx}{dt}=\frac{d}{d\tau}\left(\frac{dx}{d\tau} \frac{d \tau}{dt}\right)= \frac{d}{d\tau}\left(\frac{dx}{d\tau} \frac{1}{\gamma_p}\right)\\=a_x\frac{1}{\gamma_p}-\frac{1}{\gamma_p^2}\frac{d \gamma_p}{d\tau}\frac{dx}{d\tau}= a_x\frac{1}{\gamma_p}-\frac{1}{\gamma_p^2}\gamma_p^3\frac{v_p^2}{c^2}\frac{d }{d \tau}\frac{dx}{dt}\] (6)
From (5) and (6), we obtain,
\[\Rightarrow \frac{d }{d \tau}\left(\frac{dx}{dt}\right)\left[1+\gamma_p\frac{v_p^2}{c^2}\right]=a_x\frac{1}{\gamma_p}\]
\[\Rightarrow \frac{d }{d \tau}\left(\frac{dx}{dt}\right)=a_x\frac{1}{\gamma_p}\left[1+\gamma_p\frac{v_p^2}{c^2}\right]^{-1}\]
\[a_t=\gamma_p^2 \frac{v_p}{c^2}a_x\left[1+\gamma_p\frac{v_p^2}{c^2}\right]^{-1}\] (7)
For one dimensional motion
\[c^2a_t^2-a_x^2 \le 0\](8)
\[c^2\left[\gamma_p^2 \frac{v_p}{c^2}a_x\left[1+\gamma_p\frac{v_p^2}{c^2}\right]^{-1}\right]^2-a_x^2 \le 0\] (9)
\[\frac{1}{c^2}\gamma_p^4 v_p^2 a_x^2\left[1+\gamma_p\frac{v_p^2}{c^2}\right]^{-2}-a_x^2 \le 0\]
\[\frac{1}{c^2}\gamma_p^4 v_p^2 \left[1+\gamma_p\frac{v_p^2}{c^2}\right]^{-2} \le 1\]
\[\frac{1}{c^2}\gamma_p^2 v_p^2 \left[\frac{1}{\gamma_p}+\frac{v_p^2}{c^2}\right]^{-2} \le 1\] (10)
With v_p=v_x tending to c the left side of the last inequation given by (10)tends to infinity while the right side stays on unity.There is a breakdown.
 
Last edited:
Anamitra Palit said:
We start with the fact that the norm-square of the four acceleration vector is negative or zero.
Indeed
\[c^2=c^2\left(\frac {dt}{d \tau}\right)^2-\left(\frac {dx}{d \tau}\right)^2-\left(\frac {dy}{d \tau}\right)^2-\left(\frac {dz}{d \tau}\right)^2\] (1)
Differentiating both sides with respect to time.....
Click to expand...
Hi. No point imo doing that. I find that the first ostensibly general '4-acceleration' equation (1) above reduces to
c((dt/dτ)^2 - 1) = dx/dτ - (2),
where we conveniently choose an axis along x direction so dy/dτ = dz/dτ = 0
Since this is strictly SR regime, we have
|dx/dτ| = v - (3),
the instantaneous speed measured in the proper frame, and
dt/dτ immediately reduces to
dt/dτ = γ = 1/√(1-v²/c²) - (4)
The time dilation factor for a clock in relative motion at speed v wrt the proper frame.
Substituting (4) back into (2) gives
c(γ^2 - 1) = v - (5)
which on substituting β = v/c further reduces to
vβ/(1 - β^2) = v - (6),
finally giving
β(1 + β) = 1 - (7)
Whose only positive solution is the normalized speed
β ~ 0.618

Your (1) is therefore a very restrictive equation dealing with velocity not acceleration. That basic error propagates so pointless going any further with your derivations.
Wikipedia provides various expressions for 4-acceleration which would be worth checking against before again attempting to post anything on that here:
https://en.wikipedia.org/wiki/Four-acceleration
 
Last edited:
Q-reeus said:
Hi. No point imo doing that. I find that the first ostensibly general '4-acceleration' equation (1) above reduces to
c((dt/dτ)^2 - 1) = dx/dτ - (2),
where we conveniently choose an axis along x direction so dy/dτ = dz/dτ = 0
Since this is strictly SR regime, we have
|dx/dτ| = v - (3),
the instantaneous speed measured in the proper frame, and
dt/dτ immediately reduces to
dt/dτ = γ = 1/√(1-v²/c²) - (4)
The time dilation factor for a clock in relative motion at speed v wrt the proper frame.
Substituting (4) back into (2) gives
c(γ^2 - 1) = v - (5)
which on substituting β = v/c further reduces to
vβ/(1 - β^2) = v - (6),
finally giving
β(1 + β) = 1 - (7)
Whose only positive solution is the normalized speed
β ~ 0.618

Your (1) is therefore a very restrictive equation dealing with velocity not acceleration. That basic error propagates so pointless going any further with your derivations.
Wikipedia provides various expressions for 4-acceleration which would be worth checking against before again attempting to post anything on that here:
https://en.wikipedia.org/wiki/Four-acceleration
Click to expand...

[It might be necessary to refresh the page for viewing the formulas and the equations properly ]
For one dimensional motion [along the x axis by way of example]we have,
\[c^2=c^2 \left(\frac{dt}{d \tau}\right)^2-\left(\frac{dx}{d \tau}\right)^2\] (1)
\[c^2 \left[\left(\frac{dt}{d \tau}\right)^2-1\right]=\left(\frac{dx}{d \tau}\right)^2\] (2)
\[c^2 \left[\gamma^2-1\right]={v_{proper}}^2\] (3)
Where
\[\gamma^2=\frac{1}{1-\frac{v^2}{c^2}};v=\frac{dx}{dt};v_{proper}=\frac{dx}{d\tau}\]Instead of (5) given by Q-Reeus we have (3)
On revering the steps (5) leads us back to (1). We do not have an insane equation like (7) as provided by Q-reeus
We may also proceed in the following manner from (3)
\[c^2\left(\gamma^2-1\right)=\left(\frac{dx}{d\tau}\right)^2=\ left(\frac{dx}{dt}\frac{dt}{d\tau}\right)^2\]

\[c^2\left(\gamma^2-1\right) =v^2 \gamma^2\]

\[\gamma^2-1=\frac{v^2}{c^2}\gamma^2\]

\[\gamma^2\left[1-\frac{v^2}{c^2}\right]=1\]

We do have a valid relation,

\[\gamma^2 \frac{1}{\gamma^2}=1\]



Thus equation (1) does not have any restriction.It contains information about acceleration though in an implicit manner. Q-reeus sis expected to review his ideas
 
Last edited:
Anamitra Palit said:
[It might be necessary to refresh the page for viewing the formulas and the equations properly ]
For one dimensional motion [along the x axis by way of example]we have,
\[c^2=c^2 \left(\frac{dt}{d \tau}\right)^2-\left(\frac{dx}{d \tau}\right)^2\] (1)
\[c^2 \left[\left(\frac{dt}{d \tau}\right)^2-1\right]=\left(\frac{dx}{d \tau}\right)^2\] (2)
\[c^2 \left[\gamma^2-1\right]={v_{proper}}^2\] (3)
Where
\[\gamma^2=\frac{1}{1-\frac{v^2}{c^2}};v=\frac{dx}{dt};v_{proper}=\frac{dx}{d\tau}\]Instead of (5) given by Q-Reeus we have (3)
On revering the steps (5) leads us back to (1). We do not have an insane equation like (7) as provided by Q-reeus
We may also proceed in the following manner from (3)
\[c^2\left(\gamma^2-1\right)=\left(\frac{dx}{d\tau}\right)^2=\ left(\frac{dx}{dt}\frac{dt}{d\tau}\right)^2\]

\[c^2\left(\gamma^2-1\right) =v^2 \gamma^2\]

\[\gamma^2-1=\frac{v^2}{c^2}\gamma^2\]

\[\gamma^2\left[1-\frac{v^2}{c^2}\right]=1\]

We do have a valid relation,

\[\gamma^2 \frac{1}{\gamma^2}=1\]



Thus equation (1) does not have any restriction.It contains information about acceleration though in an implicit manner. Q-reeus sis expected to review his ideas
Click to expand...
Indeed I forgot to apply a square root to that enclosed in parentheses in LHS of my (2). This then effects my (5) which should have read
c√(γ^2 - 1) = γv - (5)
where γv is the 4-velocity spatial component measured in the particle frame, whereas -v is the particle 3-velocity measured in the lab frame.

Which then agrees with your new (3) in #7 and onward to yield the trivial equality γ^2/γ^2 = 1

It's still the case your initial (1) is purely a velocity relation with no acceleration terms present, hence cannot represent the norm squared of 4-acceleration. See post #2 here:
https://www.physicsforums.com/threads/the-norm-of-four-acceleration.588686/
The norm of 4-acceleration a in terms of coordinate 3-vectors v and a is shown there as
a·a = -γ^6a·a + γ^6[(v·a)² - (v·v)(a·a)]
Nothing like your initial (1)
 
Got it. where Anamitra begins in #1 with
"We start with the fact that the norm-square of the four acceleration vector is negative or zero. Indeed...."
.... he actually starts with an expression for 4-velocity. And then proceeds to derive the 4-acceleration as next step. Got it.
 
Four acceleration[1] in terms of coordinate time derivatives in place of proper time derivatives has been considered [ A Wikipedia link has been provided at the end of the writing.The current post addresses the issue of four acceleration in a general manner[in terms of coordinate time derivatives,considering the Wikipedia article on Four acceleration] The issues raised by Q-reeus have been discussed.Contents of my earlier posts have consequently been validated proving the assertions of Q-reeus quite invalid]:
\[A=\left(\gamma_u {\dot{\gamma}}_u c,\gamma_u^2 \vec{a}+\gamma_u {\dot{\gamma}}_u\vec{u}\right)=\left(\gamma_u^4\frac{\vec{a}.\vec{u}}{c},\gamma_u^4\left(\vec{a}+\frac{\vec{u }\times \left(\vec{u}\times \vec{a}\right)}{c^2}\right)\right)\] (1)
[The vectors u and a are time derivatives and not proper time derivatives. Dot represents time derivatives]
\[A.A= \gamma_u ^8\left( \frac{\left(\vec{a}.\vec{u}\right)^2}{c^2}- \left(\vec{a}+\frac{\vec{u }\times \left(\vec{u}\times \vec{a}\right)}{c^2}\right).\left(\vec{a}+\frac{\vec{u }\times \left(\vec{u}\times \vec{a}\right)}{c^2}\right)\right)\] (2)
\[A.A= \gamma_u ^8\left( \frac{\left(\vec{a}.\vec{u}\right)^2}{c^2}- \left(\vec{a}+\frac{\vec{u }\left(\vec{u}.\vec{a}\right)-\vec{a}u^2)}{c^2}\right).\left(\vec{a}+\frac{\vec{u }\left(\vec{u}.\vec{a}\right)-\vec{a}u^2}{c^2}\right)\right)\](3)
\[\Rightarrow A.A=\gamma_u^8\left(\frac{\left(\vec{a}.\vec{u}\right)^2}{c^2}-\left(a^2+\frac{u^2\left(\vec{u}.\vec{a}\right)^2+a^2u^4-2\left(\vec{u}.\vec{a}\right)^2u^2}{c^4}+2\frac{\left(\vec{a}.\vec{u}\right)^2-a^2u^2}{c^2}\right)\right)\]
\[\Rightarrow A.A=\gamma_u^8\left(\frac{\left(\vec{a}.\vec{u}\right)^2}{c^2}-a^2-\frac{-u^2\left(\vec{u}.\vec{a}\right)^2+a^2u^4}{c^4}-2\frac{\left(\vec{a}.\vec{u}\right)^2-a^2u^2}{c^2}\right)\]
\[\Rightarrow A.A=\gamma_u^8\left(-\frac{\left(\vec{a}.\vec{u}\right)^2}{c^2}-a^2+\frac{u^2\left(\vec{u}.\vec{a}\right)^2}{c^4}-\frac{a^2u^4}{c^4}+2\frac{a^2u^2}{c^2}\right)\]
\[\Rightarrow A.A=\gamma_u^8\left(-\frac{\left(\vec{a}.\vec{u}\right)^2}{c^2}+\frac{u^2\left(\vec{u}.\vec{a}\right)^2}{c^4}-a^2+\frac{a^2u^2}{c^2}+\frac{a^2u^2}{c^2}-\frac{a^2u^4}{c^4}\right)\]
\[\Rightarrow A.A=\gamma_u^8\left[-\frac{\left(\vec{a}.\vec{u}\right)^2}{c^2}\left(1-\frac{u^2}{c^2}\right)-a^2\left(1-\frac{u^2}{c^2}\right)+\frac{a^2u^2}{c^2}\left(1-\frac{u^2}{c^2}\right)\right]\]
\[\Rightarrow A.A=-\gamma_u^8\left(1-\frac{u^2}{c^2}\right)\left[\frac{\left(\vec{a}.\vec{u}\right)^2}{c^2}+a^2-\frac{a^2u^2}{c^2}\right]\]
\[\Rightarrow A.A=-\gamma_u^8\left(1-\frac{u^2}{c^2}\right)\left[\frac{\left(\vec{a}.\vec{u}\right)^2}{c^2}+a^2\left(1-\frac{u^2}{c^2}\right)\right]\]
\[\Rightarrow A.A=-\gamma_u^6\left[\frac{\left(\vec{a}.\vec{u}\right)^2}{c^2}+a^2\left(1-\frac{u^2}{c^2}\right)\right]\] (4)
\[\Rightarrow A.A\le 0\] (5)
The equality sign may occur if velocity[three velocity] is constant.
We may prove that
\[A=\left(c\frac{d^2t}{d\tau^2},\frac{d^2x}{d\tau^2},\frac{d^2y}{d\tau^2},\frac{d^2z}{d\tau^2}\right)=\left(\gamma_u {\dot{\gamma}}_u c,\gamma_u^2 \vec{a}+\gamma_u {\dot{\gamma}}_u\vec{u}\right)\]
justifying (1)
Indeed
\[c\frac{d^2t}{d\tau^2}=c\frac{d}{d \tau}\left(\frac{dt}{d\tau}\right)=c\frac{d\gamma_u}{d\tau}=c\frac{d\gamma_u}{dt}\frac{dt}{d\tau}=c \gamma_u {\dot{\gamma}}_u\]
\[c\frac{d^2t}{d\tau^2}=c \gamma_u {\dot{\gamma}}_u\] (6.1)
\[\frac{d^2x}{d\tau^2}=\frac{d}{d \tau}\left(\frac{dx}{d\tau}\right)=\frac{d}{dt}\left(\frac{dx}{dt}\frac{dt}{d\tau}\right)\frac{dt}{d\tau}\\=\left(\frac{d^2x}{dt^2}\gamma_u+\frac{dx}{dt}\frac{d\gamma_u}{dt}\right)\gamma_u\]
\[\frac{d^2x}{d\tau^2}=\frac{d^2x}{dt^2}\gamma_u^2+\frac{dx}{dt}{\dot{\gamma}}_u\gamma_u\]
\[\frac{d^2x}{d\tau^2}=a_x\gamma_u^2+v_x{\dot{\gamma}}_u\gamma_u\] (6.2)
In a similar manner we may show that
\[\frac{d^2y}{d\tau^2}=a_y\gamma_u^2+v_y{\dot{\gamma}}_u\gamma_u\] (6.3)
\[\frac{d^2z}{d\tau^2}=a_z\gamma_u^2+v_z{\dot{\gamma}}_u\gamma_u\] (6.4)
[a_x,a_y and a_z represent the second order time derivatives]
Therefore,
\[A=\left(c\frac{d^2t}{d\tau^2},\frac{d^2x}{d\tau^2},\frac{d^2y}{d\tau^2},\frac{d^2z}{d\tau^2}\right)\\=\left(\gamma_u {\dot{\gamma}}_u c,a_x\gamma_u^2+v_x{\dot{\gamma}}_u\gamma_u,a_y\gamma_u^2+v_y{\dot{\gamma}}_u\gamma_u,a_z\gamma_u^2+z{\dot{\gamma}}_u\gamma_u\right)\]\[\Rightarrow A=\left(\gamma_u {\dot{\gamma}}_u c,\vec{a}\gamma_u^2+\vec{u}{\dot{\gamma}}_u\gamma_u\right)\] (7)
Thus we have justified (1)
Four Velocity, General Perspectives
Differentials dx^alpha behave like contravariant tensors
Indeed,
\[d\bar {x}^\mu=\frac{\partial \bar {x}^\mu}{\partial x^\alpha}dx^\alpha\] (8.1)
\[\frac{d\bar {x}^\mu}{d\tau}=\frac{\partial \bar {x}^\mu}{\partial x^\alpha}\frac{dx^\alpha}{d\tau}\](8.2)
Equations (8.1) and (8.2) are universally valid[mathematically valid] irrespective of the presence or the absence of non zero higher order derivatives like d^2x^alpha/ tau
The tensor nature of dx^alpha or of dx^alpha/d tau is always maintained. For space time coordinates both in Special as well as in General Relativity[irrespective of the presence or absence of accelerations] we have as four velocity the quadruplet.
\[v=\left(c\frac{dt}{d\tau},\frac{dx}{d\tau},\frac{dy}{d\tau},\frac{dz}{d\tau}\right)\] (9)
\[v=\left(c\gamma_u,\frac{dx}{dt}\gamma_u,\frac{dy}{dt}\gamma_u,\frac{dz}{dt}\gamma_u\right)=\left(c\gamma_u,v_x\gamma_u,v_y\gamma_u,,v_z\gamma_u\right)=\left(c\gamma_u,\vec{u}\gamma_u\right)\](10)
[In (10) v_ i represents the coordinate time derivatives and not the proper time derivatives]
Proof of v.A=0 [ using the current relations]:
Considering equations (7) and (9) we have,
\[v.A=c^2\gamma_u^2{\dot{\gamma}_u}-\vec{u}\gamma_u\left(\vec{a}\gamma_u^2+{\dot{\gamma}_u}\gamma_u\vec{u}\right)\](11)
\[\Rightarrow v.A=c^2\gamma_u^2{\dot{\gamma}}_u-\gamma_u^3\vec{u}.\vec{a}-\gamma_u^2{\dot{\gamma}}_uu^2\]
\[\Rightarrow v.A=c^2\gamma_u^2{\dot{\gamma}}_u\left(1-\frac{u^2}{c^2}\right)-\gamma_u^3\vec{u}.\vec{a}\]
\[\Rightarrow v.A=c^2{\dot{\gamma}}_u-\gamma_u^3\vec{u}.\vec{a}\](12)
\[\Rightarrow v.A=c^2\frac{d \gamma_u}{dt}-\gamma_u^3\vec{u}.\vec{a}=c^2\gamma_u^3 u\frac{du}{dt}\frac{1}{c^2}-\gamma_u^3\vec{u}.\vec{a}\]
\[\Rightarrow v.A=\gamma_u^3 u\frac{du}{dt}-\gamma_u^3\vec{u}.\vec{a}\] (13)
Now
\[u\frac{du}{dt}=\frac{1}{2}\frac{d\left(u^2\right)}{dt}=\frac{1}{2}\frac{d \left(\vec{u}.\vec{u}\right)}{dt}=\frac{1}{2}2 \vec{u}\frac{d\vec{u}}{dt}=\vec{u}.\vec{a}\]
\[u\frac{du}{dt}=\vec{u}.\vec{a}\] (14)
From (11) and (12) we have
\[v.A=0\] (15)
Reference
1. Wikipedia, Four Acceleration,
https://en.wikipedia.org/wiki/Four-acceleration#:~:text=Four-acceleration in inertial coordinates,-In inertial coordinates&text=Geometrically, four-acceleration is a curvature vector of a worldline.&text=When the four-force is,reduces to the geodesic equation.
Accessed on 10th March,2021
 
Last edited:
You must log in or register to reply here.
Back
Top