# Anamitra Palit

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1. ## Vector Subspaces

We consider a linear vector space V of dimension n. W is a proper subspace of V.We take a vector 'e' belonging to V-W and N vectors y_i belonging to W;i=1,2,3…N;N>>n the dimension of V. All y_i cannot obviously be independent the number N being greater than the dimension of V the parent vector space; k of the y_i vectors are considered to be linearly independent where k is the dimension of W. The rest of the y_i are linear combinations of these k, basic y_i vectors of W. We consider sums αi=e+yi;i=1,2…N (1) Now each alpha_i=e+y_i belongs to V-W. We prove it as follows If possible let alpha_i belong to W. We have e=αi−yi=αi+(−yi) (2) Both alpha_i and -y_i belong to W. Therefore their sum ‘e’ should belong to W . This contradicts our postulate that e belongs to V-W. Therefore each alpha_i belongs to V-W. Next we consider the equation Σi(ciαi)=0 (3) ⇒Σi=Ni=1(ci(e+yi))=0 ⇒eΣi=Ni=1ci=−Σi=Ni=1ciyi (4) The right side of (4) belongs to W while the left side belongs to V-W. If the left side belonged to W them (1/Sigma c_i)(Sigma c_i)e=e would belong to W which is not the case. |The right side being a linear combination of vectors from W belongs to W.The only solution to avoid this predicament would be to assume Sigma c_i on the left side of (4) to be zero: that each side of (4) represents the null vector. We cannot have all c_i=0 [individually]in an exclusive manner since that would make the space N dimensional , in view of (3), [N is much greater than n, the dimension of the parent vector space V]. Equations Σi=Ni=1ci=0 (3.1) Σi=Ni=1ciyi=0 (3.2) From (3.1) cN=−c1−c2−c3…..−cN−1 (4) Considering (3.2) with (4) we have, yN=c1c1+c2+c3…..+cN−1y1+c2c1+c2+c3…..+cN−1y2+…..+cN−1c1+c2+c3…..+cN−1yN−1 (5.1) yN=a1y1+a2y2+……aN−1yN−1 (5.2) Where, ai=cic1+c2+…+cN−1 (5.3) From (5.3) we have the identity a1+a2+…+aN−1=1 (6) But the N(>>N) vectors were chosen arbitrarily. Equation (5.2) should not come under the constraint of equation (6).We could have chosen y_N in the form of (5.2) in a manner that (6) is violated.
2. ## Accelerating Particles in Spacial Relativity

First we consider 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 [We may prove by alternative techniques that a.a<=0] 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) $\gamma_p=\frac{1}{\sqrt{1-\frac{v_p^2}{c^2}}}=\frac{1}{\sqrt{1-\frac{v_x^2}{c^2}}}$ 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.
3. ## On Four Velocity and Four Momentum

Forum expert Markus Hanke declined believing in the relation v1.v2>=c^2 that stems from the reversed Cauchy Schwarz inequality.In case he happens to believe in the stated relation my discussion has made a substantial contributed to the forum and it[my discussion] is far from being of a speculative nature. Kino seems to have accepted this relation v1.v2>=c^2 [The relation v.v=c^2[gamma^2-1+1/gamma^2] is not a valid one . Nevertheless v1.v2>=c^2 is a valid formula] I have corrected a loose statement made by the forum expert. |his indeed has a non speculative contribution. A writing that locates a contradiction in the exiting theory[formal theory] The norm of the acceleration four vector is negative[or zero , for uniform motion].There are different methods of proving the stated fact. One is presented here. We start with the metric $c^2d\tau ^2=c^2dt^2-dx^2-dy^2-dz^2$ (1.1) $c^2=c^2 \left(\frac{dt}{d\tau}\right)^2-\left(\frac{dx}{d\tau}\right)^2-\left(\frac{dy}{d\tau}\right)^2-\left(\frac{dz}{d\tau}\right)^2$ (1.2) Differentiating both sides of (1.2) with respect to proper time tau, we obtain, $c^2 \frac{dt}{d\tau}\frac{d^2 t}{d \tau^2}- \frac{dx}{d\tau}\frac{d^2 x}{d \tau^2} - \frac{dx}{d\tau}\frac{d^2y}{d\tau^2}-\frac{dz}{d\tau}\frac{d^2 z}{d \tau^2} =0$(3) Next we transform to an inertial frame of reference where the particle is momentarily[instantaneously] at rest [maintaining the acceleration of the original frame] Equation (3) reduces to : $c^2 \frac{dt}{d\tau}\frac{d^2 t}{d \tau^2}\frac{d^2 t}{d\tau^2}=0$ Since v.v=c^2 and v_x=v_y=v_z=0 implying gamma=1 that is dt/d tau=1,we have $c^2 \frac{d^2 t}{d \tau^2}=0$(4) From (4) we infer that a.a<=0 in the reference frame we have transformed to. From the invariance of the dot product we have $a.a \le 0$ (5) for all inertial frames of reference. We consider one dimensional motion in the x-x’ direction for which a_yand a_z are zero [implying from the Lorentz transformations a’_y=a’_z=0 ]. The reference frames are translating in the x-x’ direction with a uniform speed v.The accelerating particle is also moving in the same direction. a.a<=0 implies $c^2a_t^2-a_x^2\le 0$ (6) Again $a_t=\frac{d^2 t}{d \tau^2}=\frac{d\gamma}{d \tau}=\gamma^3 v_x a_x \frac {1}{c^2}$ where, $\gamma=\frac{1}{\sqrt{1-\frac{v_x^2}{c^2}}}$ We have, $a_t= \gamma^3 v_x a_x \frac {1}{c^2}$(7) From (6) and (7) we have $\frac{1}{c^2} \gamma^6 v_x^2 a_x^2\le a_x^2$(8) For non zero a_x, $\frac{1}{c^2} \gamma^6 v_x ^2\le 1$(9) For v_x tending to c the last equation fails. There is no reason to believe that by discussion is of a speculative nature.Requesting the moderator to restore it to its normal status.
4. ## On Four Velocity and Four Momentum

The acceleration four vector is indeed space like or null[for uniform motion] The norm of the acceleration four vector as suggested by Kino is negative[or zero , for uniform motion].There are different methods of proving the stated fact. One is presented here as a prelude to the point referred to finally. We start with the metric $c^2d\tau ^2=c^2dt^2-dx^2-dy^2-dz^2$ (1.1) $c^2=c^2 \left(\frac{dt}{d\tau}\right)^2-\left(\frac{dx}{d\tau}\right)^2-\left(\frac{dy}{d\tau}\right)^2-\left(\frac{dz}{d\tau}\right)^2$ (1.2) Differentiating both sides of (1.2) with respect to proper time tau, we obtain, $c^2 \frac{dt}{d\tau}\frac{d^2 t}{d \tau^2}- \frac{dx}{d\tau}\frac{d^2 x}{d \tau^2} - \frac{dx}{d\tau}\frac{d^2y}{d\tau^2}-\frac{dz}{d\tau}\frac{d^2 z}{d \tau^2} =0$(3) Next we transform to an inertial frame of reference where the particle is momentarily[instantaneously] at rest [maintaining the acceleration of the original frame] Equation (3) reduces to : $c^2 \frac{dt}{d\tau}\frac{d^2 t}{d \tau^2}\frac{d^2 t}{d\tau^2}=0$ Since v.v=c^2 and v_x=v_y=v_z=0 implying gamma=1 that is dt/d tau=1,we have $c^2 \frac{d^2 t}{d \tau^2}=0$(4) From (4) we infer that a.a<=0 in the reference frame we have transformed to. From the invariance of the dot product we have $a.a \le 0$ (5) for all inertial frames of reference. We consider one dimensional motion in the x-x’ direction for which a_yand a_z are zero [implying from the Lorentz transformations a’_y=a’_z=0 ]. The reference frames are translating in the x-x’ direction with a uniform speed v.The accelerating particle is also moving in the same direction. a.a<=0 implies $c^2a_t^2-a_x^2\le 0$ (6) Again $a_t=\frac{d^2 t}{d \tau^2}=\frac{d\gamma}{d \tau}=\gamma^3 v_x a_x \frac {1}{c^2}$ where, $\gamma=\frac{1}{\sqrt{1-\frac{v_x^2}{c^2}}}$ We have, $a_t= \gamma^3 v_x a_x \frac {1}{c^2}$(7) From (6) and (7) we have $\frac{1}{c^2} \gamma^6 v_x^2 a_x^2\le a_x^2$(8) For non zero a_x, $\frac{1}{c^2} \gamma^6 v_x ^2\le 1$(9) For v_x tending to c the last equation fails.
5. ## On Four Velocity and Four Momentum

The section of my second last post from after equation (17.2) and before the section " The Sum and Difference of Two Proper Velocities " has been revised We restate (13),(14),(15) ..(17.1) and then proceed.. $v.v= c^2$(13) $v.ka =0$(14) Therefore $v.\left(v-ka\right) = c^2$ (15) We could expect from the last equation $\left(v-ka\right)$ to be a proper velocity. Then norm^2=c^2. But we can vary k and make the norm different from c^2 so that v-ka is not a proper velocity. Let $\left||v-ka\right||=\bar c^2 \ne c^2$ (16) We adjust the value of k so that norm^2 of v-ka is positive. This permits the application of the reversed Cauchy Schwarz inequality From (a) by applying reversed Cauchy Schwarz we have for v and v-ka $v.\left(v-ka\right) \ge cc'$(17.1) $\Rightarrow c^2\ge cc'\Rightarrow c\ge c'$ That is c'=||v-ka||<=c (A) If ka could be a proper velocity then ||v-ka||^2<=2c^2 [we have had this earlier from the second last post] from The Sum and Difference of Two Proper Speeds. $||v-ka||^2\le 2c^2$ (B) if ka is a proper speed We show that ka can indeed be a proper speed: Norm square of the acceleration vector $c^2a_t^2-a_x^2-a_y^2-a_z^2=\bar c^2$ c bar has the unit of acceleration $m^2\left(c^2a_t^2-a_x^2-a_y^2-a_z^2\right)=m^2\bar c^2$ m in the above equation has the dimension of time and we make the magnitude of m unity [in the concerned units]. Thus we have $c^2a_t^2-a_x^2-a_y^2-a_z^2=\bar c^2$ Next we multiply both sides of the above equation by dimensionless k^2 so that k* c bar is equal to c. Now we have $c^2k^2a_t^2-k^2a_x^2-k^2a_y^2-k^2a_z^2=k^2\bar c^2$ $c^2\left(k a_t\right)^2-\left(k a_x\right)^2-\left(k a_y\right)|^2 -\left(k a_z\right)^2=c^2$ Therefore ka is a candidate for a proper velocity.But (A) and (B) contradict each other
6. ## On Four Velocity and Four Momentum

Equation (7) of the last post has been rewritten[it could not be parsed in the last post and there was no time left for editing]: $v_1.v_2=cv_{1t}cv_{2t}-v_{1x}v_{2x}-v_{1y}v_{2y}-v_{1z}v_{2z}\ge \\ \sqrt{c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2}\sqrt{c^2v_{2t}^2-v_{2x}^2-v_{2y}^2-v_{2z}^2}=c^2$ For writing (3.3) in the last post we have the following considerations For X>=0, X>=X cos theta Therefore, a1 b1+X>=a1b1+X cos theta or, a1b1-X cos theta>=a1 b1-X X:product of two square roots, a positive number For X>=0 $X\ge X \cos \theta$ $a_1 b_1+X\ge a_1b_1+X\cos\theta$ $\Rightarrow a_1b_1-X\cos\theta\ge a_1b_1-X$ $X=\sqrt{a_2^2+a_3^2+.....+a_n^2}\sqrt{b_2^2+b_3^2+.....+b_n^2}$ NB: $v_1.v_2\le -c^2$ will not apply since $v_1.v_1=c^2>0$ [We can make v2=v1]
7. ## On Four Velocity and Four Momentum

The current form of the article in Latex [It might be necessary to refresh the page for viewing the ormulas and the equations] Mathematical Theorem If $\left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\right)\ge 0$ then either $\left(a_1b_1-a_2b_2\right)\ge \sqrt{a_1^2-a_2^2}\sqrt{b_1^2-b_2^2}$(1.1) or $\left(a_1b_1-a_2b_2\right)\le -\sqrt{a_1^2-a_2^2}\sqrt{b_1^2-b_2^2}$(1.2) Proof $\left(a_1b_1-a_2b_2\right)^2-\left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\right)=\left(a_1b_2-a_2b_1\right)^2$ $\left(a_1b_1-a_2b_2\right)^2-\left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\right)\ge 0$ $\left(a_1b_1-a_2b_2\right)^2\ge \left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\right)$ The last inequality implies (1.1) or (1.2)[subject to the initial criterion] Next we prove the reversed Cauchy Schwarz inequality: If $\left(a_1^2-a_2^2-a_3^2-......-a_n^2\right)\left(b_1^2-b_2^2-b_3^2-....-b_n^2\right)\ge 0$ Then $\left(a_1b_1-a_2b_2-a_3b_3-....-a_nb_n \right)\ge \\ \sqrt{a_1^2-a_2^2-a_3^2-.....a_n^2}\sqrt{b_1^2-b_2^2-b_3^2-....-b_n^2}$ (2.1) or $\left(a_1b_1-a_2b_2-a_3b_3-....-a_nb_n\right)\le \\ -\sqrt{a_1^2-a_2^2-a_3^2-....a_n^2}\sqrt{b_1^2-b_2^2-b_3^2-....-b_n^2}$ (2.2) Proof: Applying the Cauchy Schwarz inequality we have, $\left(a_2b_2+a_3b_3+....+a_nb_n\right)^2\ge \left(a_2^2+a_3^2+....+a_n^2\right)\left(a_2^2+a_3^2+....+a_n^2\right)$ $\left[\frac{a_2 b_2+a_3b_3+...a_nb_n}{\sqrt{a_2^2+a_3^2+a_4^2+...+a_n^2}\sqrt{b_2^2+b_3^2+b_4^2+...+b_n^2}}\right]^2\le 1$ $\Rightarrow -1 \le \frac{a_2 b_2+a_3b_3+...a_nb_n}{\sqrt{a_2^2+a_3^2+a_4^2+...+a_n^2}\sqrt{b_2^2+b_3^2+b_4^2+...+b_n^2}}\le 1$ $\Rightarrow \frac{a_2 b_2+a_3b_3+...a_nb_n}{\sqrt{a_2^2+a_3^2+a_4^2+...+a_n^2}\sqrt{b_2^2+b_3^2+b_4^2+...+b_n^2}}=\cos \theta$ (3.1) $\Rightarrow a_2b_2+a_3b_3+...a_nb_n=\\ \sqrt{a_2^2+a_3^2+a_4^2+...+a_n^2}\sqrt{b_2^2+b_3^2+b_4^2+...+b_n^2}\cos \theta$(3.2) Since the square roots to the right of the last equation are positive quantities, $\Rightarrow a_1b_1-a_2b_2-a_3b_3-...-a_nb_n=\\ a_1 b_1-\sqrt{a_2^2+a_3^2+a_4^2+...+a_n^2}\sqrt{b_2^2+b_3^2+b_4^2+...+b_n^2}\cos \theta$(3.2) But $a_1 b_1-\sqrt{a_2^2+a_3^2+a_4^2+...+a_n^2}\sqrt{b_2^2+b_3^2+b_4^2+...+b_n^2}\cos \theta\ge \\ a_1 b_1-\sqrt{a_2^2+a_3^2+a_4^2+...+a_n^2}\sqrt{b_2^2+b_3^2+b_4^2+...+b_n^2}$(3.3) Using our mathematical theorem, $a_1 b_1-\sqrt{a_2^2+a_3^2+a_4^2+...+a_n^2}\sqrt{b_2^2+b_3^2+b_4^2+...+b_n^2} \ge \\ \sqrt{a_1^2-a_2^2-a_3^2-a_4^2-...-a_n^2}\sqrt{b_1^2-b_2^2-b_3^2-b_4^2-...-b_n^2}$|(4) provided $\left(a_1^2-a_2^2-a_3^2-a_4^2-...-a_n^2\right)\left(b_1^2-b_2^2-b_3^2-b_4^2-...-b_n^2\right)\ge 0$ Therefore, subject to the above criterion, $\Rightarrow \left( a_1 b_1-a_2b_2-a_3b_3-...-a_nb_n\right)^2\ge\left(a_1^2-a_2^2-a_3^2-a_4^2-...-a_n^2\right)\left(b_1^2-b_2^2-b_3^2-b_4^2-...-b_n^2\right)$(5) $\Rightarrow \left( a_1 b1-a_2b_2-a_3b_3-...-a_nb_n\right)\ge \\ \sqrt{a_1^2-a_2^2-a_3^2-a_4^2-...-a_n^2}\sqrt {b_1^2-b_2^2-b_3^2-b_4^2-...-b_n^2}$ or $\Rightarrow \left( a_1 b1-a_2b_2-a_3b_3-...-a_nb_n\right)\le \\ -\sqrt{a_1^2-a_2^2-a_3^2-a_4^2-...-a_n^2}\sqrt {b_1^2-b_2^2-b_3^2-b_4^2-...-b_n^2}$(6) Wikipedia link for the reversed Cauchy Schwarz inequality[in relation to relativity] https://en.wikipedia.org/wiki/Minkowski_space#Norm_and_reversed_Cauchy_inequality Keeping in the mind $\left(c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2\right)\left(c^2v_{2t}^2-v_{2x}^2-v_{2y}^2-v_{2z}^2\right)\ge 0$ we have by the reversed Cauchy Schwarz inequality $v_1.v_2=cv_{1t}cv_{2t}-v_{1x}v_{2x}-v_{1y}v_{2y}-v_{1z}v_{2z}\ge \\ \sqrt{c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2}|\sqrt(c^2v_{2t}^2-v_{2x}^2-v_{2y}^2-v_{2z}^2}=c^2$(7) $v_1.v_2\ge c^2$ (8) Metric $c^2d\tau^2=c^2dt^2-dx^2-dy^2-dz^2$(9) $\Rightarrow c^2=c^2\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dx}{d\tau}\right)^2-\left(\frac{dy}{d\tau}\right)^2-\left(\frac{dz}{d\tau}\right)^2$(10) Differentiating both sides of the above with respect to proper time we obtain, $c^2\frac{dt}{d\tau}\frac{d^2t}{d\tau^2}-\frac{dx}{d\tau}\frac{d^2x}{d\tau^2}-\frac{dy}{d\tau}\frac{d^2y}{d\tau^2}-\frac{dz}{d\tau}\frac{d^2z}{d\tau^2}=0$(11) $\Rightarrow v.a=0$(12) Next we consider the following two equations $v.v= c^2$(13) $v.ka =0$(14) Therefore $v.\left(v-ka\right) = c^2$ (15) We could expect from the last equation $\left(v-ka\right)$ to be a proper velocity. Then norm^2=c^2. But we can vary k and make the norm different from c^2 so that v-ka is not a proper velocity. Let $\left||v-ka\right||=\bar c^2 \ne c^2$ (16) We adjust the value of k so that norm^2 of v-ka is positive but diffeent from c^2. This permits the application of the reversed Cauchy Schwarz inequality From (a) by applying reversed Cauchy Schwarz we have for v and v-ka $v.\left(v-ka\right) \ge cc'$(17.1) or $v.\left(v-ka\right) \le -cc'$ (17.2) But c' is adjustable. Equation (A) fails The Sum and Difference of Two Proper Speeds $c^2=c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2$(18.1) $c^2=c^2v_{2t}^2-v_{2x}^2-v_{2y}^2-v_{2z}^2$ (18.2) Adding the last two equations $2c^2=c^2\left(v_{1t}^2+v_{2t}^2\right)-\left(v_{1x}^2+v_{2x}^2\right)-\left(v_{1y}^2+v_{2y}^2\right)-\left(v_{1z}^2+v_{2z}^2\right)$ $2c^2=c^2\left(v_{1t}+v_{1t}\right)^2-\left(v_{1x}+v_{2x}\right)^2-\left(v_{1y}+v_{2y}\right)^2-\left(v_{1z}+v_{2z}\right)^2-2v_1.v_2$(19) $2c^2+2v_1.v_2=c^2\left(v_{1t}+v_{2t}\right)^2-\left(v_{1x}+v_{2x}\right)^2-\left(v_{1y}+v_{2y}\right)^2-\left(v_{1z}+v_{2z}\right)^2$ $c^2\left(v_{1t}+v_{2t}\right)^2-\left(v_{1x}+v_{2x}\right)^2-\left(v_{1y}+v_{2y}\right)^2-\left(v_{1z}+v_{2z}\right)^2 \ge 4c^2$(20) If v1 and v2 are four velocities then $\left ||v_1+v_2 \right||^2\ge 4c^2 \ne c^2$(21) Thus the sum of two proper velocities is not a proper velocity since the norm fails to match. In a vector space it is not necessary that all vectors should have the same norm. Nevertheless in the physical or in the intuitive sense it is quite uncanny! The Difference of Two Proper Speeds: $2c^2=c^2\left(v_{1t}^2-v_{2t}^2\right)-\left(v_{1x}^2-v_{2x}^2\right)-\left(v_{1y}^2-v_{2y}^2\right)-\left(v_{1z}^2-v_{2z}^2\right)$ $2c^2=c^2\left(v_{1t}^2+v_{2t}^2\right)-\left(v_{1x}^2+v_{2x}^2\right)-\left(v_{1y}^2+v_{2y}^2\right)-\left(v_{1z}^2+v_{2z}^2\right)-2v_2.v_2$ $2c^2=c^2\left(v_{1t}-v_{1t}\right)^2-\left(v_{1x}-v_{2x}\right)^2-\left(v_{1y}-v_{2y}\right)^2-\left(v_{1z}-v_{2z}\right)^2+2v_1.v_2-2v_2.v_2$ $2c^2-2v_1.v_2+2v_2.v_2=c^2\left(v_{1t}-v_{2t}\right)^2-\left(v_{1x}-v_{2x}\right)^2-\left(v_{1y}-v_{2y}\right)^2-\left(v_{1z}-v_{2z}\right)^2$ $c^2\left(v_{1t}-v_{2t}\right)^2-\left(v_{1x}-v_{2x}\right)^2-\left(v_{1y}-v_{2y}\right)^2-\left(v_{1z}-v_{2z}\right)^2 \le 2c^2$(22) If v1 and v2 are four velocities then $\left ||v_1-v_2 \right||^2\le 2c^2 \ne c^2$(23) Reversing the sign of v2 in (19) we obtain (23) We may solve the following three equations $c^2=c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2$(24.1) $c^2=c^2v_{2t}^2-v_{2x}^2-v_{2y}^2-v_{2z}^2$(24.2)) $c^2\left(v_{1t}+v_{2t}\right)^2-\left(v_{1x}+v_{2x}\right)^2-\left(v_{1y}+v_{2y}\right)^2-\left(v_{1z}+v_{2z}\right)^2$(24.3) We have from the last three equations, $v_1.v_2 \le -\frac{1}{2}c^2$(24.4) Equation (24.3)assumes the sum of two proper speeds to be a proper speed. The interesting point is that solutions do exist for these equation: some of the components are not real.If the three equations are viewed mathematically regardless of any physics they imply some sort of a contradiction.equation (24.4) stands in opposition to equation (8) We may repeat the mathematical theorem and the derivation of the Cauchy |Schwarz inequality considering ai and bj to be complex numbers with $a_1 b_1-a_2 b_2-a_3b_3-....-a_nb_n$ and $\left(a_1^2-a_2^2-a_3^2.......-a_n^2\right)\left(b_1^2-b_2^2-b_3^2.......-b_n^2\right)$ as real with $\left(a_1^2-a_2^2-a_3^2.......-a_n^2\right)\left(b_1^2-b_2^2-b_3^2.......-b_n^2\right)\ge 0$ Existence of solutions to (21.1),(24.2) and (24.3) or equivalently those to (24.1),(24.2) and (24.4) ,irrespective of the nature of the solutions being real or complex, are indicative of a subtle issue of contradiction revealed by the simultaneous validity of equations (8) and (24.4).[We may solve (24.1),(24.2) and (24.4) directly ] One has to think of equations(24.1),(24.2) and (24.3) in the mathematical sense regardless of the physics involved with them.
8. ## O the Riemann Curvature Tensor

$X_{\alpha\beta\gamma\delta}\left(g^{\alpha\mu}g^{\beta\nu}+ g^{\beta\mu}g^{\alpha\nu}\right)=0$ The coefficient matrix is independent of gamma and delta. For the same alpha and beta various gamma and delta should produce the same value for the corresponding component ,that is, X_{alpha beta gamma delta}=X_{alpha beta gamma’ delta’} If the coefficient matrix is non zero(or zero) for some (gamma,delta ) pair it should remain non zero (or zero) for other (gamma, delta) pairs.
9. ## O the Riemann Curvature Tensor

1)General perspective always pertain to the special cases.Whatever ensues from: $X_{\alpha\beta\gamma\delta}\left(g^{\alpha\mu}g^{\beta\nu}+g^{\alpha\nu}g^{\beta\mu}\right)=0$ will also apply to the Riemann tensor. For constant delta and gamma there are 16 variables of the type X_{alpha beta gamma delta} for four values of each alpha and beta..These are treated as the unknowns. There are 16 equations for the four values of each of mu and nu. We have sixteen linear homogeneous equations with sixteen variables[unknowns].If the determinant of the coefficient matrix is non zero then each X_{alpha beta gamma delta}=0. If you consider the Riemann tensor in place of X the component values will be zero[subject to the determinant of the coefficient matrix being non zero] . This idea is quite evident in my last post in the discussion. [gamma and delta have been held constant everywhere in the comment] 2)Two proper velocity vectors will not add up to a proper velocity since the norm square of the sum in general will be different from c^2: gre ater than c^2 3)The comment made by Kino on the "Norm of the Four Velocity Vector" is quite misleading. He is not specific with the error. He is searching for it as I understand.He will definitely report it at his earliest convenience.I hope so!
10. ## O the Riemann Curvature Tensor

It is not a correct point.General perspectives cannot be violated in the special cases.Why don't you go through my last post in this discussion?
11. ## O the Riemann Curvature Tensor

[It might be necessary to refresh the page for proper viewing] Thanking Kino for his first comment.[posted on Saturday,3:43 pm] It is true that the product of a symmetric and antisymmetric tensor is always zero. Nevertheless we may consider the following: $R_{\alpha\beta\gamma\delta}\left(g^{\alpha\mu}g^{\beta\nu}+ g^{\beta\mu}g^{\alpha\nu}\right)=0$(1) [A significant aspect of (1) is that it holds for any arbitrary(mu,nu) pair] Keeping gamma and delta constant[and we maintain so throughout this post] we vary alpha and beta for each equation in (1). In order to obtain new equations of the type (1) we vary mu and nu There are 16 unknowns in R_αβγδ for the four alpha and the four beta while we have 16 equations[linear homogeneous equations ] in that mu and nu take on four values of each. Unless the determinant of the coefficient matrix is zero the solutions to R_αβγδ are trivial in their nature. This cannot be violated subject to a coefficient matrix with non zero determinant] We have $R_{\alpha\beta\gamma\delta}=0$ (2) [The null tensor is always antisymmetric[ asides being symmetric]] We may write (1) in the form $\left(R_{\alpha\beta\gamma\delta}+R_{\beta\alpha\gamma\delta}\right)\left(g^{\alpha\mu}g^{\beta\nu}+ g^{\beta\mu}g^{\alpha\nu}\right)=0$(3) We have eight variables of the form $R_{\alpha\beta\gamma\delta}+R_{\beta\alpha\gamma\delta}$ while the number of equations continue to remain sixteen. Gamma and delta have been maintained constant throughout the writing as mentioned earlier.
12. ## The Enigma of the Tensors

[One my have to refresh the page for viewing the formulas and the equations] We consider the transformation of the rank two contravariant tensor: $\bar A^{\mu \nu}=\frac{\partial \bar x^{\mu}}{\partial x^\alpha}\frac{\partial \bar x^{\nu}}{\partial x^\beta}A^{\alpha\beta}$(1) Inverse transformation[for non singular transformations] $A^{\alpha \beta}=\frac{\partial x^\alpha}{\partial \bar x^\mu}\frac{\partial x^{\beta}}{\partial \bar x^\nu}\bar A^{\mu\nu}$(2) For the diagonal components[alpha=beta] $A^{\alpha \alpha}=\frac{\partial x^\alpha}{\partial \bar x^\mu}\frac{\partial x^{\alpha}}{\partial \bar x^\nu}\bar A^{\mu\nu}$ (3) We consider the situation where the off diagonal components of A[alpha not equal to beta] are all zero The diagonal elements of A-bar are given by $\bar A^{\mu\mu}=\left(\frac{\partial \bar x^\mu}{\partial x^\alpha}\right)^2 A^{\alpha\alpha}$ (4) For off diagonal elements of A-bar are given by ' $\bar A^{\mu\nu}=\frac{\partial \bar x^\mu}{\partial x^\alpha} \frac{\partial \bar x^\mu}{\partial x^\alpha} A^{\alpha\alpha}$ (4’) Subject to the situation that the off diagonal elements of A are all zero we consider the following three cases: Case 1. Assume A-bar^ mu nu=0for all μ≠ν for all barred reference frames[the general condition that off diagonal elements of A are zero continues to hold] we have $\frac{\partial \bar x^\mu}{\partial x^\alpha} \frac{\partial \bar x^\mu}{\partial x^\alpha} A^{\alpha\alpha}=0$ (4’’) We have (4'') irrespective of the transformation elements[all reference frames being considered]. From (4’’) A-bar^mu mu=0.Again from (4’) and (4’’) |A-bar ^mu nu becomes zero for μ≠ν. Then the tensor becomes null Case 2 Off diagonal elements are non zero for all A-bar[that the off diagonal elements of A are all zero continues to hold] $\bar A^{\mu\mu}=\left(\frac{\partial \bar x^\mu}{\partial x^\alpha}\right)\frac{\partial x^\alpha}{\partial \bar x^\rho} \frac{\partial x^\alpha}{\partial \bar x^\sigma} A^{\rho\sigma}$ $\left(\frac{\partial \bar x^\mu}{\partial x^\alpha}\right)\frac{\partial x^\alpha}{\partial \bar x^\rho} \frac{\partial x^\alpha}{\partial \bar x^\sigma} A^{\rho\sigma}=\delta^{\mu}_{\rho} \delta^{\mu}_{\sigma}$ $\frac{\partial \bar x^\rho}{\partial x^\beta}\frac{\partial \bar x^\sigma}{\partial x^\gamma}\left(\frac{\partial \bar x^\mu}{\partial x^\alpha}\right)\frac{\partial x^\alpha}{\partial \bar x^\rho} \frac{\partial x^\alpha}{\partial \bar x^\sigma} A^{\rho\sigma}=\frac{\partial \bar x^\rho}{\partial x^\beta}\frac{\partial \bar x^\sigma}{\partial x^\gamma}\delta^{\mu}_{\rho} \delta^{\mu}_{\sigma}$ $\Rightarrow\left(\frac{\partial \bar x^\mu}{\partial x^\alpha}\right)^2\delta^\alpha_\beta \delta^\alpha_\gamma=\frac{\partial \bar x^\mu} {\partial x^\beta}\frac{\partial \bar x^\mu}{\partial x ^\gamma}$ For β≠γ we have from (5) $\frac{\partial \bar x^\mu}{\partial x ^\beta}\frac{\partial \bar x^\mu}{\partial x ^\gamma}=0$(6) $\Rightarrow\frac{\partial \bar x^\mu}{\partial x^\beta}=0$or $\frac{\partial \bar x^\mu}{\partial x ^\beta}=0$ (7) $\Rightarrow \bar A^{\mu\nu}=0\Rightarrow A^{\alpha\beta}=0$ (8) Equations represented by (7) and (8)are not true! We do have an enigma of a persistent nature. [In fact (8) implies that the metric tensor is the null tensor;the Riemann tensor and the Ricci tensor are null tensors; the Ricci scalar is zero valued.] Case 3 For A as per our initial postulation the off diagonal elements of A are all zero. For A-bar there we assume the existence of at least one mu ,nu pair in each barred frame such that A-bar^mu nu=0.These mu nu pairs may not be identical for all barred frames. Since we may have an infinitude of reference frames catering to our assumption it follows that for some specific (μ,ν)pair also A-bar mu mu is zero in an infinitely many frames of reference $\bar A^{\mu\nu}=\frac{\partial \bar x^\mu}{\partial x^\alpha} \frac{\partial \bar x^\nu}{\partial x^\alpha}A^{\alpha\alpha}$ Since A^m mu=0 for a specific pair in infinitely many frames,we have, $\frac{\partial \bar x^\mu}{\partial x^\alpha} \frac{\partial \bar x^\nu}{\partial x^\alpha}A^{\alpha\alpha}=0$(7) For the same mu nu pair (7) we have an infinitude of equations like (7) whee the transformation elements are distinct. We have included infinitely many barred frames in equation (7) $\Rightarrow A^{\alpha\alpha}=0\Rightarrow A=0$ (8) Case 1 is an instance of case 3 Points to Observe: 1. Anti symmetric tensors transform to anti symmetric tensors in all frames of reference. Noe the null tensor is an antisymmetic tensor asides being a symmetric one.An arbitrary non zero[non null] tensor may be expressed as the sum of a symmetric an an antisymmetric tensor. Our analysis might be considered for an arbitrary tensor which is neither symmetric or antisymmetric.It reduces the null tensor making both the symmetric and the antisymmetric parts zero. 2.If a tensor has a component of identical value in all reference frames it has to be the null tensor. 3. Let us represent (1) in a standard matrix form $\bar A=MAM^T$(9) where M is the transformation matrix For an arbitrary M even if M is non singular in nature, we will not necessarily have both A and A-bar as diagonal[both with the off diagonal elements as zero].If one of A or A-bar is diagonal with the other non diagonal then A ,A-bar both become null tensors as we have already seen! One may test the situation using a diagonal tensor with A and an arbitrary square matrix for M For an arbitrary transformation matrix ,a diagonal tensor A does not necessarily produce a diagonal A-bar. Only by specific valid choice of the space time transformation can we have both sides of (10) or (1) for that matter, as diagonal[components zero for alpha not equal to beta]. An arbitrary[non singular ] transformation will produce from a diagonal tensor a non diagonal. It is always possible to have infinitely many transformation for which the off diagonal elements are non zero[diagonal elements of the tensor in some reference frame are assumed to be non zero]
13. ## O the Riemann Curvature Tensor

Link to file on the Google Drive https://drive.google.com/file/d/1C2-ru6uuDIw9u_e4HQ1bdwa01oKysQ-B/view?usp=sharing Material in Latex[It might be necessary to refresh the page for viewing the formulas and the equations] The paper establishes mathematically that the Riemann Tensor is a zero tensor Riemann Curvature |Tensor $R^{\mu\nu}_{\quad\gamma\delta}=g^{\alpha\mu}g^{\beta\nu}R_{\alpha\beta\gamma\delta}$ (1) Interchanging the dummy indices alpha and beta we have, $R^{\mu\nu}_{\quad\gamma\delta}=g^{\beta\mu}g^{\alpha\nu}R_{\beta\alpha\gamma\delta}$(2) Therefore, $g^{\alpha\mu}g^{\beta\nu}R_{\alpha\beta\gamma\delta}=g^{\beta\mu}g^{\alpha\nu}R_{\beta\alpha\gamma\delta}$ (3) But $R_{\beta\alpha\mu\nu}=-R_{\alpha\beta\mu\nu}$ (4) Thus we have $g^{\alpha\mu}g^{\beta\nu}R_{\alpha\beta\gamma\delta}=-g^{\beta\mu}g^{\alpha\nu}R_{\alpha\beta\gamma\delta}$ (5) $R_{\alpha\beta\gamma\delta}\left[g^{\alpha\mu}g^{\beta\nu}+g^{\beta\mu}g^{\alpha\nu}\right]=0$ (6) With (6) alpha and beta are dummy indices;others are free indices.Treating the Riemann tensor components as variables[unknowns], there are sixteen of them for the sixteen alpha beta combinations any given gamma , delta and mu, nu combinations.Each equation has gamma,delta,mu and nu as constant quantities. In total we have 256 equations[four values for each gamma,delta,mu and nu].Since these equations are of homogeneous nature, $R_{\alpha\beta\gamma\delta}=0$ (7) In the orthogonal system of coordinates we obtain from (6) for distince mu and nu $R_{\mu\nu\gamma\delta}\left[g^{\mu\mu}g^{\nu\nu}+g^{\nu\mu}g^{\mu \nu}\right]=0$ (8) In (8) we have considered different values of mu and nu as well as different values for gamma and delta. One should also take note of the fact that there is no summation on mu and on nu. $R_{\mu\nu\gamma\delta}g^{\mu\mu}g^{\nu\nu}=0$ (9) In (8) also there is no summation on mu and on nu. Since g_mu mu not equal to 0 and g_nu nu not equal to zero we have for unequal mu ,nu and unequal gamma delta, $R_{\mu\nu\gamma\delta}=0$ (10) If the Riemann tensor is a zero tensor ten the Ricci tensor is also a zero tensor.The Ricci scala becomes a zero valued scalar. Riemann Curvature.pdf
14. ## Inferences from the General Relativity Metric

(t,x,y,z) could represent coordinates of four dimensional space in a general manner; they could be Cartesian,Spherical or any other system. Equation (1) of the initial post represents the most general type of a General Relativity metric in the orthogonal system: $c^2d\tau^2= c^2g_{00} d t^2-g_{11} dx^2-g_{22} dy^2-g_{33} dz^2$ [Inadvertently,with the string post of the discussion, after copy pasting I forgot to change the suffix with g_ii to appropriate values.]
15. ## Inferences from the General Relativity Metric

[One may have to refresh the page for viewing the formulas and the equations] 1. In our discussion we have considered time like separations at a point so that d tau^2>0 2. Towards the end of the last post..For arbitrary K,l and M we observe, point (2) has to be ignored 3. We had $cdt=kdr=ld\theta=md \phi$.All three components in the last line have the dimensions of length. k is dimensionless while l and m have the dimensions of length. Dividing the Schwarzschild metric by (cdt)^2 we obtain $\left(\frac {d\tau}{dt}\right)^2=\left(1-\frac{2Gm}{c^2 r}\right)-\frac{1}{k^2}\left(1-\frac{2Gm}{c^2 r}\right)^{-1}-\frac{1}{l^2} r^2-\frac{1}{m^2}r^2 \sin^2 \theta$ In the last equation we may set k=1 ,l=m=1 unit of length. Therefore, $\left(1-\frac{2Gm}{c^2 r}\right)\ge \frac{1}{k^2}\left(1-\frac{2Gm}{c^2 r}\right)^{-1}+\frac{1}{l^2}r^2+\frac{1}{m^2}r^2 \sin\theta$ In the last equation has to hold for any k,l and m for which dtau^2 is positive.[as a particular instance we may set k=1 ,l=m=1 unit of length] and for arbitrary r beyond the event horizon:dt,dx,dy and dz are not present but their rlative sizes are present. For large values of 'r' the metric automatically becomes space like unless we make dtheta and dphi extraordinarily small with respect to dr and dt Plain text versions of the two stated formulas: [dtau/dt]^2=[1-2Gm/c^2r]-(1/k^2)[1-2Gm/c^2r]^(-)-(1/l^2)r^2-(1/m^2)r^2sin^2 theta Since time like separations are being considered on (+,-,-,-) [1-2Gm/c^2r]>(1/k^2)[1-2Gm/c^2r]^(-)+(1/l^2)r^2+(1/m^2)r^2sin^2 theta the above inequation will not hold for large r. For a general type of a metric given the space time location we have to be careful with the relative sizes of the coordinate intervals in order to have a time like or a space like interval.
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