sidharath

suppose the same mass is attached to the moving and stationary spring balances, will the springs be stretched to the same extent in both the cases

values of physical constants like gravitational constant G , permittivity etc depends on dimensions of length, mass , time etc therefore the magnitude of the constant is expected to be not same in the moving and stationary frame So .these are not constants because magnitude changes rewlativily

Dear studiot you express in your post 18 that as the pendulum moves towards mean position change in potential energy is negative,kinetic energy acquired by pendulum is equal to change in potential energy ,therefore if change in PE is negative KE will be negative but KE is always positive. If the sign of potential energy is negative as pendulum moves towards meanm position the change in PE will be positive hence KE will be positive,no rule will be violated

certain mass is attached to stationary spring balance, next the spring balance is subjected to very high upward velocity Will the spring balance show high reading due to increase in relative mass as a result of high velocity ?

According to stud idiot when the particle is moving away or towards mean position KE decteases and increases respectively.Let us see what it leads to . In SHO T+U=constant so dU=-dT when the particle is moving away from mean position KE decreases hence dT is negative so dU is positive it is possible only if expression for PE is +1/2kx^2 . If the particle is moving towards mean position KE increases .dT is positive .dU is negative which is possible only if U=-1/2kx^2. it can be concluded that there is change in sign of PE

I request all of you intelligent persons that you specifically point out the error in my calculation . The calculations are based on standard relations and conventions.If the final result got by using the acknowledged relations is wrong where i have gone wrong in using the relations.. .. Are the standard relations used by me wrong ? Dear Studiot s' result is based on graphs .while my result is based on analytically calculations. Again i beg your help to pin point where lies the error in my calculations. Let this matter be conclusively settled . I shall be thankful to you

Let us take the case of sign of potential energy of SHO.There is two way motion of the particle in SHO therefore the sign of potential energy is expected to change with direction of motion of particle.When the particle is moving away work is being done on the particle while when it i moving towards mean position particle is doing work hence sign of potential energy should be reversed. The follo wing discussion conclusively proves it the basic formula to calculate potential energy is dU=-w ,U is the potential energy and w is work. Work is calculated by applying dot product formula .w=scalar value of force x scalar value of displacement x cos(angle between direction of force and displacement) when the particle is moving away from mean position w==-kxdx because angle is 18o dU=kxdx on integration U=1/2kx^2 because potential energy at mean position is 0. when the particle is moving towards mean position w=kxdx because angle is 0 so dU=-kxdx on integration U= -1/2KX^2 The above discussion proves that sign of potential is negative when particle moves towards mean position. This basic fact is not mentione in text books Please let me know where i have gone wrong .

the basic relation between potential energy and work energy is given below dU=-w Uis potential energy and w is work . Suppose the particle is moving away from mean position. work w can be calculated by applying dot product of force and distance. if x is the displacement w=modulus of force x modulus of distance x cos (angle between direction of force and distance) when the particle is moving away from mean position angle is 180 .therefore dU=kx .dx kx is the value of force and xis displacement because cos180=-1. On integrating the above relation U=1/2 k x^2 which shows that value of potential energy is positive. if the particle is moving away from mean position the angle between force and displacement is o the value of cos 0=1 which leads to dU=--kx dx on integrating U=-1/2kx^2 which shows that sign of potential energy changes to negative sign. The change in sign with change in direction is not mentioned in textbooks Please express your opinion

please devote some time to answer the following queries related to the topic 1.there is increase in mass of the accelerated electron in cyclotron and adjustment has to be made for it.., in the case under study mass along with the spring and cabin is accelerated hence there is real extension whether you observe it from inside or outside the cabin 2if one pair of electron and positron moving with very high speed in opposite directions is annihilated while another pair moving slowly in opposite directions is annihilated the frequency of photons liberated in the former case will be more than latter case due to more increase in mass in the former case 3The force formula when the mass is moving with high speed is not equal to mass into acceleration d

suppose in thought experiment there are two cabins , one cabin is plaed at rest in gravitational field while the second cabin is accelarated in direction from bottom to ceiling with acceleration g.. In both the cabins elastic string is attached to he ceiling.To the ewn of the spring in both the cabins certain mass is attached..The spring will be stretched in both the cabins but with difference that the extent of stretching will not change with time in the cabin placed in gravitational field while the extent of stretching will go on increasig with time in the accelerated cabin . It is so because the spring is attached to cabin hence its velocity and the velocity of the mass attached to it will go on increasig with time because cabin is accelewrated..According to special theory of relativity the increase in velocity will increase the mass which will result in more stretching of the accelerated spring. Special theory of relativity is applicable to accelerated frame though with the difference that instantaneous velocity is to be considered.The above observation shows that the two frames are not equivalent but equivalence principle of general relativity states that the two cabins or frames are equivalent .

i have applied theory of relativity to arrive at the conclusion that if the ball is placed at floor of the accelerated box the mass o r energy or momentum of the ball goes on increasing due ti increase in velocity hence with time due to increase in mass greater force is needed with time to pick it up from floor (mg).I shall be thankful to you if you let me know why theory of relativity is not applicable . If there is no increase in mass , where does the energy which is associated with increase in velocity goes? Please devote some time to convey your opinion related to another thought similar to the first . Suppose electron and proton are annihilated in accelerated box and box at rest , will the energy of photons emitted from accelerated box and box or frame at rest be same .?

i am sorry that i wrote the title of the new thread as --general relativity . The title of the new thread should have been as - Equivalence Principle and i request you to discuss the new thread as such. i am eagerly waiting for your valuable thoughts so that my doubts could be removed because i approached you after going through some terxts according to special theory of relativity the ball in the moving frame or box will acquire more and more mass or energy with time because acceleration leads to increase in velocity which leads increase in mass or energy . The impact momentum depends on mass as well as velocity hence impact force increases with time in accelerated frame or box as compared to stationary frame or it can also be said that more and more force is needed to pick the ball from floor as time increases.Moreover the force formula according to special theory of relativity involves velocity in the denominator hence force changes with time actually it increases with increases in velocity because force relation is of such a form

In thought experiment a box with negligible mass isolated in space is subjected to acceleration directed in direction from floor towards ceiling.A box of certain mass when dropped from ceiling will strike the floor with certain impact force.When the experiment is repeated after sometime the ball will strike the floor with more impact force due to acceleration of box the initial velocity of ball as it starts from ceiling will be more as compared to the previous case hence according to special theory of relativity initial mass or energy or momentum of ball will be more which will lead to more impact force . The balls thrown at different time intervals will strike the floor with different impact forces. If the above experiment is repreatad in box held stationary in uniform gravitational field associated with acceleration g the ball will always strike the floor with same impact force independent of time interval..The above observation leads to conclusion that the the accelerated box or frame and the stationary frame or box placed in uniform gravitational field are not equivalent .The general theory of relativity is based on equivalence principle which here seems to be not correct .Please let me know if my conclusion is correct

thank you very much indeed for the reply If the same experiment is performed in lift which is falling down freely with acceleration g or moving upwards with acceleration g will there be compressing and stretching of the spring?Suppose the lift is not experiencing any external gravitational field , will the tension in the spring be equal to mg where m is the mass attached to the end of spring?

suppose in thought experiment a box with negligible mass is isolated from any other mass . An elastic spring is attached to the ceiling of the box while other end carries certain mass . The box is uniformly accelerated alternatively in direction from lower end of the spring towards ceiling and then in direction from ceiling towards lower end of spring .Will there be any tension in the spring in both the cases?

When electron and positron are annihilated in the moving carriage of train the energy liberated is more than when the pair is annihilated on platform which shows that there is increase in mass on motion because it is mass that is annihilated during pair interaction

The energy of mass whether at rest or in motion is product of mass and square of velocity of light.The energy of moving mass is more than that of mass at rest which is possible only if mass increases with velocity because in energy relation velocity of light is constant. It can be said that increase in mass is hybrid of mass and energy so can be named as mattenergy

if photon is associated with gravity it leads to the conclusion that photon is hybrid of two types of highly unstable particles where one is associated with pure electromagnetic field while other is carrier of pure gravity. The particles are named by me as electromagnaton and mattenergon.These particles exist in combined state only.Considering the novel nature of photon the properties of photon can be easily understood and photon is no longer an enigma as it once was was to Einstern

when photon splits into positron and electron .the pair produced is expected to be associated with gravitational field which indicates that in addition to electromagnetic field photon is also carrier of gravitational field

thank you very much studidiot for the trouble you have taken in sending detailed reply. You have mentioned correctly that raindrops strike the umbrella vertically whether it is moving or stationary. I was confused because in some sub standard text it was mentioned that when the umbrella is moving the velocity of the falling drops is the resultant of velocities of drops and that of walker acting at 90 angle hence the rain drops strike the umbrella in oblique direction as a consequence of which umbrella should be held in oblique direction for maximum protection

Thank you very much indeed ophiolite captainpanic studidiot for your reply.i shall be highly thankful to you if you let me know whether the rain drops strike the surface of the umbrella in vertical or oblique direction when the walker is moving very briskly. I am a little confused

The rain is falling down vertically and the person holding umbrella is moving with very quick pace. To avoid getting wet should the umbrella held in vertical or oblique direction.

as proved in my last mail as particle moves towards mean position dU=dT. suppose is negative u--1/2x^2 dT=--Kxdx which shows as expected that when x decreases T increases. if u is positive as particle moves towards position U=1/2Kx^2 dT=+kx dx which shows that as x decreases t decreases which is contrary to reality .hence potenteial energy of particle moving towards mean position cannot be positive. apart from the direct derivation of potential potential energy as particle moves away or towards mean position by using basic relaton du =-Fdx=-w where the sign of potential energy as moves towards mean position is found out to be negative. indirect results using the same basic relation also prove my contention.If contention is to be refuted it should be based on fundamentals otherwise you are free to give your opinion sorry there was a little error in above calculation/ As the particle moves towards mean position through distance dx work done w= -2kdx therefore du=-F dx=-w=2kdx=dT hence du=dT The rest of matter is the same . i am again feeling very sorry for minor unintentional lapse. again minor mistake please add, as particle moves from position x to x-dx towards mean position work w has been calculated for this shift

i am writing very simple and conclusive reply to support m contention that sign of potential energy particle moving away from mean position is positive while when it moves away potential sign is negative The basic relation ia dU=-dF.dx.When the particle moves opposite to restoring force work done is negative dU=W=-dT There is fall in kinetic energy because of positive work dU=-dT U=-T When the particle moves towards mean position in the restoring force direction work done is positive dU=-W=dT dU-dT orU=T IT is so because of liberation of energy there is increase in kinetic energy. suppose particle with maximum potential +E starts moving towards mean position .at certain point potential energy is reduced to U Change in potential energy IS U-E is converted to kinetic energy U-E=T or E=U-T WHICH IS WRONG. SWANSOT can shout here that E-0 MIDWAY it is concluded that particle with positive potential energy while moving towards mean position gives wrong result. if particle with maximum negative energy-E MOVES towards mean position and U IS POTENTIAL ENERGY at certain point change in potential energy U-(-E)=U+E-T E=T-U THE sign of potential U is negative E=T+U^ U=-U^ SO CORRECT RESULT IS GOT . numerical value of potential is used There is now no doubt

I am giving final reply in support of my contention which based upon fundamentals which are not of my own. I stand by the conclusions reached. I am amazed that even elementary things are not clear. In an S.H.O while the particle moves from mean position to extreme position kinetic energy is being changed to potential energy while reverse is the case when particle moves from extreme to mean position. Suppose u^ is the potential energy released. This potential energy is converted to t^ kinetic energy. Suppose particle is moving from extreme position towards mean position it has got potential energy = - 1/2(kx^2). When particle further moves from x to y position potential energy released u^ = U (final) - U (initial) = 1/2(k)(x^2 - y^2), where x > y. Therefore energy released is positive T^ = u^, where u^, therefore as expected kinetic energy increases at the expense of potential energy as particle moves towards mean position. If potential energy of the particle as it moves toward mean position the final result is t^ = 1/2(k)(y^2 - x^2), where y < x, therefore t^ is negative, which indicates that as particle moves towards mean position there is falling kinetic energy which is contrary to the fact. The above treatment conclusively proves without any doubt that correct result is got when the potential energy of the particle moving toward mean position is taken as negative. To get the correct results when the particle moves away from mean position, potential energy has to be taken as positive. Swansont has pointed out that at certain distance potential energy is taken as negative total energy E will be 0, but the following simple observation shows that even with negative potential energy it is not possible. Suppose particle moves from extreme position toward mean position with maximum potential energy - E. At a certain distance potential energy is U, sign negative is included in U. Change in potential energy to U - (- E) = U + E. The released potential energy is changed to kinetic energy T, where T = U + E, or E = T - U. Thus U = - U^, where U^ is positive. Therefore, E = E + U^, both the term on R.H.S are positive. Therefore, E can never be zero even with negative potential energy. I request to contradict my contention in such a manner that replies are based upon true fundamental facts so that you may prove me wrong, but I emphatically I am not wrong at all. It is the ignorance about basic facts that your replies are based and your are ought negate my conclusion.